Sabar's triangle

DefinitionSabar's triangle#✓fully proved

For natural numbers \(n\) and \(i\), the entry \(s(n, i)\) of Sabar's triangle is the integer

\[s(n, i) = \sum_{j=i+1}^{n} (-1)^{n-j} \binom{j}{i}.\]

In particular \(s(n, i) = 0\) for \(i \ge n\), the empty sum.

In the source, \(s(n, i)\) counts certain grid paths; for \(0 \le i \le n - 1\) these numbers form OEIS A341091.

public def sabarNumber (n i : ℕ) : ℤ :=
  ∑ j ∈ Finset.Ioc i n, (-1) ^ (n - j) * (j.choose i : ℤ)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Section 2 (definition of s(n, i)).

Used by 8

Used byFine numbers from Sabar's triangle · Harmonic numbers from Sabar's triangle · Central entries of Sabar's triangle and Fine numbers · Central entries of Sabar's triangle and Whitt's numbers · A recurrence for Sabar's triangle in small cases · Rows 5 and 10 of Sabar's triangle · Reciprocal binomial sums from Sabar's triangle · The harmonic numbers as an alternating sum of Sabar numbers

TheoremRows 5 and 10 of Sabar's triangle#✓fully proved

The entries \(s(5, i)\) of Sabar's triangle for \(i = 0, 1, \dots, 4\) are \(1, 2, 7, 6, 5\), and the entries \(s(10, i)\) for \(i = 0, 1, \dots, 9\) are \(0, 6, 24, 71, 129, 167, 147, 92, 36, 10\).

These are rows \(n = 5\) and \(n = 10\) of the source's Table 1.

public theorem sabarNumber_rows_five_ten :
    (List.range 5).map (sabarNumber 5) = [1, 2, 7, 6, 5] ∧
      (List.range 10).map (sabarNumber 10) = [0, 6, 24, 71, 129, 167, 147, 92, 36, 10]

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Table 1.

Uses 1

Statement usesSabar's triangle

TheoremA recurrence for Sabar's triangle in small cases#✓fully proved

For all natural numbers \(n\) and \(i\) with \(i < n < 8\), the entries of Sabar's triangle \(s\) satisfy \(s(n + 2, i + 1) = s(n, i + 1) + \binom{n+1}{i}\).

This is the source's third basic identity \(s(n, i) = s(n-2, i) + \binom{n-1}{i-1}\), stated here only in these cases. It fails for \(i = n\) and \(i = n + 1\) (for \(i \ge n + 2\) both sides are \(0\)): for example \(s(2, 1) = 2\) but \(s(0, 1) + \binom{1}{0} = 1\).

public theorem sabarNumber_recurrence_small :
    ∀ n < 8, ∀ i < n,
      sabarNumber (n + 2) (i + 1) = sabarNumber n (i + 1) + ((n + 1).choose i : ℤ)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Section 2 (identities following from the definition).

Uses 1

Statement usesSabar's triangle

TheoremHarmonic numbers from Sabar's triangle#●ready to prove

For every natural number \(n \ge 2\), the harmonic number \(H_n = \sum_{k=1}^{n} \frac{1}{k}\) and the entries \(s(n-1, i)\) of Sabar's triangle satisfy

\[H_n = \frac{1}{n} \sum_{i=0}^{n-1} \binom{n-1}{i}^{-1} + \sum_{i=0}^{n-2} \frac{(-1)^{n-i} \, s(n-1, i)}{n-i}.\]

public theorem_wanted harmonic_eq_sum_inv_choose_add_sum_sabarNumber (n : ℕ) (hn : 2 ≤ n) :
    harmonic n =
      (1 / (n : ℚ)) * ∑ i ∈ Finset.range n, 1 / ((n - 1).choose i : ℚ) +
        ∑ i ∈ Finset.range (n - 1), (-1 : ℚ) ^ (n - i) * sabarNumber (n - 1) i / ((n - i : ℕ) : ℚ)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Theorem 2.

Uses 1

Statement usesSabar's triangle

TheoremReciprocal binomial sums from Sabar's triangle#●ready to prove

For every natural number \(n \ge 2\), the entries \(s(n-1, i)\) of Sabar's triangle satisfy

\[\frac{1}{n+1} \sum_{i=0}^{n} \binom{n}{i}^{-1} = 1 + \sum_{i=0}^{n-2} \frac{(-1)^{n-i-1} \, s(n-1, i)}{(n-i)(n+1-i)}.\]

This is the source's Corollary 1 with the sign used in its proof. As printed, with \((-1)^{n-i}\), it fails at \(n = 2\): the left side is \(5/6\) but \(1 + s(1, 0)/(2 \cdot 3) = 7/6\).

public theorem_wanted sum_inv_choose_succ_eq_one_add_sum_sabarNumber (n : ℕ) (hn : 2 ≤ n) :
    (1 / ((n + 1 : ℕ) : ℚ)) * ∑ i ∈ Finset.range (n + 1), 1 / (n.choose i : ℚ) =
      1 + ∑ i ∈ Finset.range (n - 1),
        (-1 : ℚ) ^ (n - i - 1) * sabarNumber (n - 1) i / (((n - i : ℕ) : ℚ) * ((n + 1 - i : ℕ) : ℚ))

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Corollary 1.

Uses 1

Statement usesSabar's triangle

TheoremThe harmonic numbers as an alternating sum of Sabar numbers#●ready to prove

For every natural number \(n \ge 2\), writing \(s(m, i)\) for Sabar's triangle and \(H_n = \sum_{j=1}^{n} 1/j\) for the \(n\)-th harmonic number, in the rational numbers

\[\sum_{i=0}^{2n-2} \frac{(-1)^i \, s(2n-1, i)}{i+1} = H_n.\]

public theorem_wanted sum_sabarNumber_div_succ_eq_harmonic (n : ℕ) (hn : 2 ≤ n) :
    ∑ i ∈ Finset.range (2 * n - 1), (-1 : ℚ) ^ i * sabarNumber (2 * n - 1) i / ((i + 1 : ℕ) : ℚ) =
      harmonic n

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Theorem 3.

Uses 1

Statement usesSabar's triangle

TheoremFine numbers from Sabar's triangle#●ready to prove

For every natural number \(n \ge 2\), the Fine number \(F(n)\) is the difference of two entries of row \(2n\) of Sabar's triangle \(s\): \(F(n) = s(2n, n+1) - s(2n, n-1)\).

public theorem_wanted fineNumber_eq_sabarNumber_sub (n : ℕ) (hn : 2 ≤ n) :
    fineNumber n = (sabarNumber (2 * n) (n + 1) - sabarNumber (2 * n) (n - 1) : ℤ)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Theorem 4.

Uses 2

Statement usesFine numbers · Sabar's triangle

TheoremSabar's alternating binomial sum equal to \(n\)#●ready to prove

For every natural number \(n \ge 2\), in the rational numbers,

\[\sum_{k=1}^{n} (-1)^{k+1} \left(1 - \frac{2k}{n+1} + \frac{n}{k+1}\right) \binom{n+k}{k} = n.\]

public theorem_wanted sum_neg_one_pow_mul_choose_eq_self (n : ℕ) (hn : 2 ≤ n) :
    ∑ k ∈ Finset.Icc 1 n, (-1 : ℚ) ^ (k + 1) *
        (1 - 2 * (k : ℚ) / (n + 1) + (n : ℚ) / (k + 1)) * ((n + k).choose k : ℚ) = n

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Corollary 2.

TheoremFine numbers as an alternating binomial sum#●ready to prove

For every natural number \(n \ge 2\), the Fine number \(F(n)\) satisfies

\[F(n) = \frac{1}{2} (-1)^n + \frac{1}{2} \sum_{k=0}^{n} (-1)^{n-k} \left(1 - \frac{n}{k+1}\right) \binom{n+k}{k}.\]

public theorem_wanted fineNumber_eq_half_neg_one_pow_add_sum (n : ℕ) (hn : 2 ≤ n) :
    fineNumber n = (1 / 2) * (-1 : ℚ) ^ n +
      (1 / 2) * ∑ k ∈ Finset.range (n + 1),
        (-1 : ℚ) ^ (n - k) * (1 - (n : ℚ) / (k + 1)) * ((n + k).choose k : ℚ)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Corollary 3.

Uses 1

Statement usesFine numbers

TheoremCentral entries of Sabar's triangle and Fine numbers#●ready to prove

Let \(F(m)\) denote the Fine numbers and \(C_m\) the Catalan numbers. For every natural number \(n \ge 1\), the entries of Sabar's triangle \(s\) satisfy

\[s(2n, n) = \frac{1}{6} \binom{2n+2}{n+1} + \frac{2}{3} F(n) + (-1)^{n+1},\]

\[s(2n, n) = \frac{1}{3} (2n+1) C_n + \frac{2}{3} F(n) + (-1)^{n+1},\]

\[s(2n, n) = \frac{2}{3} \binom{2n}{n} - \frac{1}{3} F(n-1) + (-1)^{n+1},\]

and

\[s(2n-1, n) = \frac{2}{3} \binom{2n-1}{n} + \frac{1}{3} F(n-1) + (-1)^n.\]

public theorem_wanted sabarNumber_central_eq_fineNumber (n : ℕ) (hn : 1 ≤ n) :
    (sabarNumber (2 * n) n : ℚ) =
        (1 / 6) * ((2 * n + 2).choose (n + 1) : ℚ) + (2 / 3) * fineNumber n + (-1) ^ (n + 1) ∧
      (sabarNumber (2 * n) n : ℚ) =
        (1 / 3) * (2 * n + 1) * (catalan n : ℚ) + (2 / 3) * fineNumber n + (-1) ^ (n + 1) ∧
      (sabarNumber (2 * n) n : ℚ) =
        (2 / 3) * ((2 * n).choose n : ℚ) - (1 / 3) * fineNumber (n - 1) + (-1) ^ (n + 1) ∧
      (sabarNumber (2 * n - 1) n : ℚ) =
        (2 / 3) * ((2 * n - 1).choose n : ℚ) + (1 / 3) * fineNumber (n - 1) + (-1) ^ n

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Theorem 5.

Uses 2

Statement usesFine numbers · Sabar's triangle

TheoremConsecutive binomial sums weighted by powers of 2#●ready to prove

For a natural number \(m\), let \(V_m(2) = \sum_{k=0}^{m} \binom{m+k}{k} 2^k\). For every natural number \(n \ge 1\), \(V_n(2) + V_{n+1}(2) = 3 \cdot 2^n \binom{2n+2}{n+1}\).

public theorem_wanted sum_choose_mul_two_pow_add_succ (n : ℕ) (hn : 1 ≤ n) :
    ∑ k ∈ Finset.range (n + 1), (n + k).choose k * 2 ^ k +
        ∑ k ∈ Finset.range (n + 2), (n + 1 + k).choose k * 2 ^ k =
      3 * 2 ^ n * (2 * n + 2).choose (n + 1)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Lemma 1.

TheoremConsecutive Whitt generalized Catalan numbers at \(\alpha = 2\)#●ready to prove

For every natural number \(n \ge 1\), Whitt's generalized Catalan numbers at \(\alpha = 2\) satisfy

\[C(2, n) + C(2, n+1) = 2^{n+1} c_n,\]

where \(c_n = \frac{1}{n+1}\binom{2n}{n}\) is the \(n\)-th Catalan number.

public theorem_wanted whittGeneralizedCatalan_two_add_succ (n : ℕ) (hn : 1 ≤ n) :
    whittGeneralizedCatalan 2 n + whittGeneralizedCatalan 2 (n + 1) = 2 ^ (n + 1) * catalan n

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Theorem 6.

Uses 1

Statement usesWhitt's generalized Catalan numbers

TheoremCentral entries of Sabar's triangle and Whitt's numbers#●ready to prove

Let \(C(2, m)\) denote Whitt's generalized Catalan numbers at \(\alpha = 2\). For every natural number \(n \ge 1\), the central entry \(s(2n, n)\) of Sabar's triangle satisfies

\[s(2n, n) = \frac{2}{3} \binom{2n}{n} - \frac{1}{3} \cdot \frac{C(2, n)}{2^{n+1}} + (-1)^{n+1} + \frac{(-1)^n}{2^{n+1}}.\]

public theorem_wanted sabarNumber_central_eq_whittGeneralizedCatalan (n : ℕ) (hn : 1 ≤ n) :
    (sabarNumber (2 * n) n : ℚ) =
      (2 / 3) * ((2 * n).choose n : ℚ) - (1 / 3) * whittGeneralizedCatalan 2 n / 2 ^ (n + 1) +
        (-1) ^ (n + 1) + (-1) ^ n / 2 ^ (n + 1)

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Theorem 7.

Uses 2

Statement usesSabar's triangle · Whitt's generalized Catalan numbers

TheoremFine numbers from Whitt's generalized Catalan numbers#●ready to prove

For every natural number \(n \ge 1\), the Fine number \(F(n)\), indexed so that \(F(0) = 1\) and \(F(1) = 0\), and Whitt's generalized Catalan number \(C(2, n+1)\) at \(\alpha = 2\) satisfy

\[2^{n+2} F(n) = C(2, n+1) + 3(-1)^n.\]

public theorem_wanted two_pow_mul_fineNumber_eq_whittGeneralizedCatalan (n : ℕ) (hn : 1 ≤ n) :
    2 ^ (n + 2) * fineNumber n = whittGeneralizedCatalan 2 (n + 1) + 3 * (-1) ^ n

Sources

• Sabar Taoufik, A New Triangle of Numbers: Properties and Observations, INTEGERS 26 (2026), Article A27, Corollary 4.

Uses 2

Statement usesFine numbers · Whitt's generalized Catalan numbers

3 fully proved11 ready to prove

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