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 1-6 of 6 results.

A003161 A binomial coefficient sum.

Original entry on oeis.org

1, 1, 2, 9, 36, 190, 980, 5705, 33040, 204876, 1268568, 8209278, 53105976, 354331692, 2364239592, 16140234825, 110206067400, 765868074400, 5323547715200, 37525317999884, 264576141331216, 1886768082651816, 13458185494436592, 96906387191038334, 697931136204820336
Offset: 0

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Author

Keywords

Comments

The number of triples of standard tableaux of the same shape of height less than or equal to 2. - Mike Zabrocki, Mar 29 2007
From Peter Bala, Mar 20 2023: (Start)
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(3,n)}. For other cases see A361887 ({S(5,n)}) and A361890 ({S(7,n)}).
Gould (1974) proposed the problem of showing 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).
Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^k) == b(n*p^(k-1)) (mod p^(3*k)) holds for positive integers n and k and all primes p >= 5. (End)

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=3 of A357824.

Programs

  • Maple
    ogf := ((8*x-1)*(8*x+1)*hypergeom([1/4, 1/4],[1],64*x^2)^2/(x+1)-3*Int((16*x-5)*hypergeom([1/4, 1/4],[1],64*x^2)^2/(x+1)^2,x)+1)/(16*x);
    series(ogf,x=0,30); # Mark van Hoeij, May 06 2013
  • Mathematica
    Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^3,{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *)
  • PARI
    a(n)=sum(k=0,n\2, (binomial(n,k)-binomial(n,k-1))^3) /* Michael Somos, Jun 02 2005 */

Formula

a(n) = Sum_{k=0..n} A120730(n,k)^3. - Philippe Deléham, Oct 18 2008
G.f.: hypergeometric expression with an anti-derivative, see Maple program. - Mark van Hoeij, May 06 2013
Recurrence: n*(n+1)^3*(7*n^2 - 14*n + 3)*a(n) = - n*(7*n^5 - 112*n^4 + 206*n^3 + 8*n^2 - 125*n + 48)*a(n-1) + 16*(n-1)*(28*n^5 - 133*n^4 + 194*n^3 - 33*n^2 - 120*n + 61)*a(n-2) + 64*(n-2)^3*(n-1)*(7*n^2 - 4)*a(n-3). - Vaclav Kotesovec, Mar 06 2014
a(n) ~ 2^(3*n+9/2) / (9 * Pi^(3/2) * n^(5/2)). - Vaclav Kotesovec, Mar 06 2014
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^3. - Alois P. Heinz, Oct 17 2022

A129123 Number of 4-tuples of standard tableau with height less than or equal to 2.

Original entry on oeis.org

1, 1, 2, 17, 98, 882, 7812, 78129, 815474, 8955650, 101869508, 1194964498, 14374530436, 176681194276, 2212121332488, 28145258688369, 363177582488274, 4745064935840178, 62687665026816228, 836447728509168930, 11261240896657686660, 152847558411986548260
Offset: 0

Views

Author

Mike Zabrocki, Mar 29 2007

Keywords

Comments

Number of pairs of Dyck paths of semilength n with equal midpoint. - Alois P. Heinz, Oct 07 2022

Crossrefs

Programs

  • Magma
    [(&+[((n-2*j+1)/(n-j+1))^4*Binomial(n,j)^4: j in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Nov 08 2022
    
  • Maple
    b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
         `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
        end:
    a:= n-> add(b(n, n-2*j)^4, j=0..n/2):
    seq(a(n), n=0..21);  # Alois P. Heinz, Mar 25 2025
  • Mathematica
    Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^4, {k,0, Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Dec 16 2017 *)
  • PARI
    a(n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^4);
    
  • Python
    from math import comb
    def A129123(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**4 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025
  • SageMath
    def A129123(n): return sum(((n-2*j+1)/(n-j+1))^4*binomial(n,j)^4 for j in range((n//2)+1))
    [A129123(n) for n in range(31)] # G. C. Greubel, Nov 08 2022
    

Formula

a(n) = Sum_{k=0..n} A120730(n,k)^4. - Philippe Deléham, Oct 18 2008
From Vaclav Kotesovec, Dec 16 2017: (Start)
Recurrence: n*(n+1)^3*(15*n^2 - 34*n + 7)*a(n) = 2*n*(90*n^5 - 309*n^4 + 147*n^3 + 124*n^2 - 135*n + 35)*a(n-1) + 4*(n-1)^2*(4*n - 5)*(4*n - 3)*(15*n^2 - 4*n - 12)*a(n-2).
a(n) ~ 3* 2^(4*n - 1/2) / (Pi^(3/2) * n^(7/2)). (End)
a(n) = A357652(n) - A355481(n). - Alois P. Heinz, Oct 13 2022
a(n) = Sum_{j=0..floor(n/2)} ((n-2*j+1)/(n-j+1))^4 * binomial(n,j)^4. - G. C. Greubel, Nov 08 2022
a(n) = Sum_{k=0..n} binomial(n,k) * ( binomial(n,k) - binomial(n,k-1) )^3. - Seiichi Manyama, Mar 25 2025

A361887 a(n) = S(5,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, 33, 276, 4150, 65300, 1083425, 20965000, 399876876, 8461219032, 178642861782, 4010820554664, 90684123972156, 2130950905378152, 50560833176021025, 1231721051614138800, 30294218438009039800, 759645100717216142000, 19213764100954274616908, 493269287121905287769776
Offset: 0

Views

Author

Peter Bala, Mar 28 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(5,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 5-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

Crossrefs

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), 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) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).
Column k=5 of A357824.

Programs

  • Maple
    seq(add( ( binomial(n,k) - binomial(n,k-1) )^5, k = 0..floor(n/2)), n = 0..20);
  • Mathematica
    Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^5, {k,0,Floor[n/2]}], {n,0,20}] (* Vaclav Kotesovec, Aug 27 2023 *)
  • Python
    from math import comb
    def A361887(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**5 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) )^5.
From Alois P. Heinz, Apr 02 2023: (Start)
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^5.
a(n) = Sum_{j=0..n} A120730(n,j)^5.
a(n) = A357824(n,5). (End)
a(n) ~ 2^(5*n + 19/2) / (125 * Pi^(5/2) * n^(9/2)). - Vaclav Kotesovec, Aug 27 2023

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

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.

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

A382433 a(n) = S(6,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, 65, 794, 19722, 562692, 15105729, 553537490, 18107304842, 716747344436, 27247858130506, 1137502720488532, 47573235297987700, 2085487143991309320, 92820152112054862785, 4246321874111740074210, 197525644801830489637170, 9363425291004877645851300
Offset: 0

Views

Author

Seiichi Manyama, Mar 25 2025

Keywords

Crossrefs

Column k=6 of A357824.

Programs

  • Maple
    b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
         `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
        end:
    a:= n-> add(b(n, n-2*j)^6, j=0..n/2):
    seq(a(n), n=0..18);  # Alois P. Heinz, Mar 25 2025
  • Mathematica
    Table[Sum[Binomial[n,k] * (Binomial[n,k] - Binomial[n,k-1])^5, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Mar 25 2025 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)*(binomial(n, k)-binomial(n, k-1))^5);
    
  • Python
    from math import comb
    def A382433(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**6 for j in range((n>>1)+1)) # Chai Wah Wu, Mar 25 2025

Formula

a(n) = Sum_{k=0..floor(n/2)} A008315(n,k)^6.
a(n) = Sum_{k=0..n} A120730(n,k)^6.
a(n) = A357824(n,6).
a(n) = Sum_{k=0..n} binomial(n,k) * ( binomial(n,k) - binomial(n,k-1) )^5.
a(n) ~ 5 * 2^(6*n+4) / (3^(5/2) * Pi^(5/2) * n^(11/2)). - Vaclav Kotesovec, Mar 25 2025

A357825 Total number of n-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j = 0..floor(n/2).

Original entry on oeis.org

1, 1, 2, 9, 98, 4150, 562692, 211106945, 404883552194, 1766902576146876, 40519034229909243476, 2708397617879598970178238, 658332084097982587522119612196, 735037057881394837614680080889845116, 2030001034486747324990010196845670569155080
Offset: 0

Views

Author

Alois P. Heinz, Oct 14 2022

Keywords

Crossrefs

Main diagonal of A357824.

Programs

  • Maple
    b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
         `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
        end:
    a:= n-> add(b(n, n-2*j)^n, j=0..n/2):
    seq(a(n), n=0..15);
  • Mathematica
    Table[Sum[(Binomial[n, k]*(n - 2*k + 1)/(n - k + 1))^n, {k, 0, n/2}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 17 2022 *)

Formula

a(n) = A357824(n,n).
a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^n.
a(n) = Sum_{j=0..n} A120730(n,j)^n.
a(n) mod 2 = 1 <=> n in { A000225 }.
From Vaclav Kotesovec, Nov 17 2022: (Start)
a(n)^(1/n) ~ exp(-1/2) * 2^(n + 3/2) / (sqrt(Pi)*n).
Limit_{n->oo} a(n) / (2^(n^2 + 3*n/2) / (n^n * exp(n/2) * Pi^(n/2))) does not exist, see also graph. (End)
Conjecture: the superconguence a(2*p-1) == 1 (mod p^3) holds for all primes p >= 5 (checked up to p = 101). - Peter Bala, Mar 20 2023
Showing 1-6 of 6 results.