cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A361890 a(n) = S(7,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 1, 2, 129, 2316, 94510, 4939220, 211106945, 14879165560, 828070125876, 61472962084968, 4223017425122958, 325536754765395096, 25399546083773839692, 2059386837863675003112, 173281152533121109073025, 14789443838781868027714800, 1307994690673355979749969800
Offset: 0

Views

Author

Peter Bala, Mar 30 2023

Keywords

Comments

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. The present sequence is {S(7,n)}. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).
a(n) is the total number of 7-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2). - Alois P. Heinz, Apr 02 2023

Links

Crossrefs

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/ S(1,2*n-1) ).
Column k=7 of A357824.
Cf. A008315, A120730.

Programs

  • Maple
    seq(add( ( binomial(n,k) - binomial(n,k-1) )^7, k = 0..floor(n/2)), n = 0..20);
  • Mathematica
    Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^7, {k,0,Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Aug 27 2023 *)
  • Python
    from math import comb
def A361890(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**7 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

Formula

a(n) = Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^7.
From Alois P. Heinz, Apr 02 2023: (Start)
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^7.
a(n) = Sum_{j=0..n} A120730(n,j)^7.
a(n) = A357824(n,7). (End)
a(n) ~ 3 * 2^(7*n + 27/2) / (2401 * Pi^(7/2) * n^(13/2)). - Vaclav Kotesovec, Aug 27 2023

A361889 a(n) = S(5,2*n-1)/S(1,2*n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 11, 415, 30955, 3173626, 386672861, 52846226091, 7857161332715, 1246162831674580, 207990691516965886, 36176886727828945286, 6510211391453319830461, 1205449991704260042021490, 228686327051301858363357905, 44299708036441260810228742915, 8738765548899621077157770551275
Offset: 1

Views

Author

Peter Bala, Mar 29 2023

Keywords

Comments

Odd bisection of A361888.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5.

Examples

			Examples of supercongruences:
a(13) - a(1) = 1205449991704260042021490 - 1 = 3*(13^3)*182893338143568508879 == 0 (mod 13^3).
a(2*5) - a(2) = 207990691516965886 - 11 = (5^3)*7*237703647447961 == 0 (mod 5^3)

Links

Crossrefs

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

Programs

  • Maple
    seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^5/binomial(2*n-1,n-1), k = 0..n-1), n = 1..20);
  • Mathematica
    Table[Sum[(Binomial[2*n-1, k]-Binomial[2*n-1, k-1])^5 / Binomial[2*n-1, n-1], {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)
  • Python
    from math import comb
def A361889(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**5 for j in range(n))//comb((n<<1)-1,n-1) # Chai Wah Wu, Mar 25 2025

Formula

a(n) = 1/binomial(2*n-1,n-1) * Sum_{k = 0..n-1} ( (2*n - 2*k)/(2*n - k) * binomial(2*n-1,k) )^5 for n >= 1.
a(n) ~ 2^(8*n + 1) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025

A361892 a(n) = S(7,2*n-1)/S(1,2*n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

Original entry on oeis.org

1, 43, 9451, 6031627, 6571985126, 9140730357409, 14801600281919487, 26927918031565051915, 53804800109969394477580, 116002825041515533807200418, 266118189111094898593879923346, 642598035707739308769581970619393
Offset: 1

Views

Author

Peter Bala, Mar 30 2023

Keywords

Comments

Odd bisection of A361891.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5.

Links

Crossrefs

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ).

Programs

  • Maple
    seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^7/binomial(2*n-1,n-1), k = 0..n-1), n = 1..20);
  • Mathematica
    Table[Sum[(Binomial[2*n-1, k]-Binomial[2*n-1, k-1])^7 / Binomial[2*n-1, n-1], {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)
  • Python
    from math import comb
def A361892(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**7 for j in range(n))//comb((n<<1)-1,n-1) # Chai Wah Wu, Mar 25 2025

Formula

a(n) = 1/binomial(2*n-1,n-1) * Sum_{k = 0..n-1} ( (2*n - 2*k)/(2*n - k) * binomial(2*n-1,k) )^7 for n >= 1.
a(n) ~ 3 * 2^(12*n+1) / (2401 * Pi^3 * n^6). - Vaclav Kotesovec, Mar 24 2025

Extensions

Offset changed to 1 by Georg Fischer, Nov 20 2024
Showing 1-3 of 3 results.

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