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.

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A098436 Triangle of 3rd central factorial numbers T(n,k).

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 73, 36, 1, 1, 585, 1045, 100, 1, 1, 4681, 28800, 7445, 225, 1, 1, 37449, 782281, 505280, 35570, 441, 1, 1, 299593, 21159036, 33120201, 4951530, 130826, 784, 1, 1, 2396745, 571593565, 2140851900, 652061451, 33209946, 399738, 1296, 1
Offset: 0

Views

Author

Ralf Stephan, Sep 08 2004

Keywords

Examples

			  1;
  1,   1;
  1,   9,    1;
  1,  73,   36,   1;
  1, 585, 1045, 100, 1;
  ...
		

Crossrefs

First column is A023001, first diagonal is A000537.
Row sums are in A098437.
Replace in recurrence (k+1)^3 with k: A008277; with k^2: A008957 (note offsets).

Programs

  • Maple
    A098436 := proc(n,k)
        option remember;
        if k=0 or k = n then
            1;
        else
            (k+1)^3*procname(n-1,k)+procname(n-1,k-1) ;
        end if;
    end proc:
    seq(seq( A098436(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Jan 13 2025
  • Mathematica
    T[n_, n_] = 1;
    T[n_ /; n>=0, k_] /; 0<=k<=n := T[n, k] = (k+1)^3 T[n-1, k]+T[n-1, k-1];
    T[, ] = 0;
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 08 2022 *)

Formula

Recurrence: T(n, k) = (k+1)^3*T(n-1, k) + T(n-1, k-1), T(0, 0)=1.

A100255 Squares of pentagonal numbers: a(n) = (1/4)*n^2*(3*n-1)^2.

Original entry on oeis.org

0, 1, 25, 144, 484, 1225, 2601, 4900, 8464, 13689, 21025, 30976, 44100, 61009, 82369, 108900, 141376, 180625, 227529, 283024, 348100, 423801, 511225, 611524, 725904, 855625, 1002001, 1166400, 1350244, 1555009, 1782225, 2033476
Offset: 0

Views

Author

Ralf Stephan, Nov 13 2004

Keywords

Comments

More generally, the ordinary generating function for the squares of k-gonal numbers is x*(1 + (k^2 - 5)*x + (4*k^2 - 18*k + 19)*x^2 + (k - 3)^2*x^3)/(1 - x)^5. - Ilya Gutkovskiy, Apr 13 2016

Crossrefs

Cf. similar sequences of the squares of k-gonal numbers: A000537 (k = 3), A000583 (k = 4), this sequence (k = 5).

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 25, 144, 484}, 32] (* Ilya Gutkovskiy, Apr 13 2016 *)
    Table[(1/4) n^2 (3 n - 1)^2, {n, 0, 31}] (* Michael De Vlieger, Apr 13 2016 *)
  • PARI
    a(n) = (1/4)*n^2*(3*n-1)^2 \\ Altug Alkan, Apr 13 2016

Formula

a(n) = A000326(n)^2.
G.f.: x*(1+20*x+29*x^2+4*x^3)/(1-x)^5. - Colin Barker, Feb 14 2012
From Ilya Gutkovskiy, Apr 13 2016: (Start)
E.g.f.: x*(4 + 46*x + 48*x^2 + 9*x^3)*exp(x)/4.
a(n) = 5*a(n-1) - 10*(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 + 4*sqrt(3)*Pi - 36*log(3) + 4*psi_1(2/3), where psi_1 is the trigamma function. - Amiram Eldar, Apr 04 2025

A123094 Sum of first n 12th powers.

Original entry on oeis.org

0, 1, 4097, 535538, 17312754, 261453379, 2438235715, 16279522916, 84998999652, 367428536133, 1367428536133, 4505856912854, 13421957361110, 36720042483591, 93413954858887, 223160292749512, 504635269460168, 1087257506689929, 2244088888116105, 4457403807182266
Offset: 0

Views

Author

Zerinvary Lajos, Sep 27 2006

Keywords

Crossrefs

Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), A023002 (m=10), A123095 (m=11), this sequence (m=12), A181134 (m=13).

Programs

  • Magma
    [(&+[j^12: j in [0..n]]): j in [0..30]]; // G. C. Greubel, Jul 21 2021
  • Maple
    [seq(add(i^12, i=1..n), n=0..18)];
  • Mathematica
    Table[Sum[k^12, {k, n}], {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
    Accumulate[Range[0,30]^12]  (* Harvey P. Dale, Apr 26 2011 *)
  • Python
    A123094_list, m = [0], [479001600, -2634508800, 6187104000, -8083152000, 6411968640, -3162075840, 953029440, -165528000, 14676024, -519156, 4094, -1, 0 , 0]
    for _ in range(10**2):
        for i in range(13):
            m[i+1]+= m[i]
        A123094_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Sage
    [bernoulli_polynomial(n,13)/13 for n in range(1, 30)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = n*A123095(n) - Sum_{i=0..n-1} A123095(i). - Bruno Berselli, Apr 27 2010
a(n) = n * (n+1) * (2*n+1) * (105*n^10 +525*n^9 +525*n^8 -1050*n^7 -1190*n^6 +2310*n^5 +1420*n^4 -3285*n^3 -287*n^2 +2073*n -691)/2730. - Bruno Berselli, Oct 03 2010
a(n) = (-1)*Sum_{j=1..12} j*Stirling1(n+1,n+1-j)*Stirling2(n+12-j,n). - Mircea Merca, Jan 25 2014

A123095 Sum of first n 11th powers.

Original entry on oeis.org

0, 1, 2049, 179196, 4373500, 53201625, 415998681, 2393325424, 10983260016, 42364319625, 142364319625, 427675990236, 1170684360924, 2962844754961, 7012409924625, 15662165784000, 33254351828416, 67526248136049, 131794658215281, 248284917113500, 453084917113500
Offset: 0

Views

Author

Zerinvary Lajos, Sep 27 2006

Keywords

Crossrefs

Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), A023002 (m=10), this sequence (m=11), A123094 (m=12), A181134 (m=13).

Programs

  • Magma
    [(&+[j^11: j in [0..n]]): n in [0..30]]; // G. C. Greubel, Jul 21 2021
    
  • Maple
    [seq(add(i^11, i=1..n), n=0..20)];
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^11 od: seq(a[n], n=0..13); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    Table[Sum[k^11, {k, n}], {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
    Accumulate[Range[0,20]^11] (* Harvey P. Dale, Sep 17 2021 *)
  • Python
    A123095_list, m = [0], [39916800, -199584000, 419126400, -479001600, 322494480, -129230640, 29607600, -3498000, 171006, -2046, 1, 0 , 0]
    for _ in range(10**2):
        for i in range(12):
            m[i+1]+= m[i]
        A123095_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014
    
  • Sage
    [(bernoulli_polynomial(n+1, 12) - bernoulli(12))/12  for n in (0..30)] # G. C. Greubel, Jul 21 2021

Formula

a(n) = n*A023002(n) - Sum_{i=0..n-1} A023002(i). - Bruno Berselli, Apr 27 2010
a(n) = n^2*(n+1)^2*(2*n^8 +8*n^7 +4*n^6 -16*n^5 -5*n^4 +26*n^3 -3*n^2 -20*n +10)/24. - Bruno Berselli, Oct 03 2010
G.f.: x*(x^10 +2036*x^9 +152637*x^8 +2203488*x^7 +9738114*x^6 +15724248*x^5 +9738114*x^4 +2203488*x^3 +152637*x^2 +2036*x +1)/(1-x)^13. - Colin Barker, May 27 2012
a(n) = (-1)*Sum_{j=1..11} j*Stirling1(n+1,n+1-j)*Stirling2(n+11-j,n). - Mircea Merca, Jan 25 2014
a(n) = 1728*A006542(n+2)^2 + 216*A288876(n-2) + 96*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, May 25 2024

A127777 A127773 * A002260 as infinite lower triangular matrices.

Original entry on oeis.org

1, 3, 6, 6, 12, 18, 10, 20, 30, 40, 15, 30, 45, 60, 75, 21, 42, 63, 84, 105, 126, 28, 56, 84, 112, 140, 168, 196, 36, 72, 108, 144, 180, 216, 252, 288, 45, 90, 135, 180, 225, 270, 315, 360, 405, 55, 110, 165, 220, 275, 330, 385, 440, 495, 550, 66, 132, 198
Offset: 1

Views

Author

Gary W. Adamson, Jan 28 2007

Keywords

Comments

Triangular number transform of A002260.
Swapped order of the factors: A002260 * A127773 = A127778.

Examples

			First few rows of the triangle:
   1;
   3,  6;
   6, 12, 18;
  10, 20, 30, 40;
  15, 30, 45, 60, 75;
  ...
		

Crossrefs

Cf. A000217, A127773, A000537 (row sums), A127778.

Programs

  • Maple
    T(n,k):=piecewise(k<=n,k*binomial(n+1,n-1),nMircea Merca, Apr 11 2012
  • Mathematica
    Table[k*Binomial[n+1,n-1],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Oct 26 2016 *)

Formula

T(n,k) = k*binomial(n+1,n-1) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+2-k,n-i), 1 <= k <= n. - Mircea Merca, Apr 11 2012

Extensions

More terms from Harvey P. Dale, Oct 26 2016

A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 9, 10, 5, 1, 1, 6, 10, 16, 15, 6, 1, 1, 5, 18, 20, 25, 21, 7, 1, 1, 8, 15, 40, 35, 36, 28, 8, 1, 1, 9, 27, 35, 75, 56, 49, 36, 9, 1
Offset: 1

Views

Author

Alford Arnold, Mar 29 2007

Keywords

Comments

The array can be constructed by beginning with A007318 (Pascal's triangle) placing each diagonal on a prime row. The other rows are filled in by mapping the prime factorization of the row number to the known sequences on the prime rows and multiplying term by term.

Examples

			Row six begins 1 6 18 40 75 126 ... because rows two and three are
1 2 3 4 5 6 ...
1 3 6 10 15 21 ...
The array begins
1 1 1 1 1 1 1 1 1 A000012
1 2 3 4 5 6 7 8 9 A000027
1 3 6 10 15 21 28 36 45 A000217
1 4 9 16 25 36 49 64 81 A000290
1 4 10 20 35 56 84 120 165 A000292
1 6 18 40 75 126 196 288 405 A002411
1 5 15 35 70 126 210 330 495 A000332
1 8 27 64 125 216 343 512 729 A000578
1 9 36 100 225 441 784 1296 2025 A000537
1 8 30 80 175 336 588 960 1485 A002417
1 6 21 56 126 252 462 792 1287 A000389
1 12 54 160 375 756 1372 2304 3645 A019582
1 7 28 84 210 462 924 1716 3003 A000579
1 10 45 140 350 756 1470 2640 4455 A027800
1 12 60 200 525 1176 2352 4320 7425 A004302
1 16 81 256 625 1296 2401 4096 6561 A000583
1 8 36 120 330 792 1716 3432 6435 A000580
1 18 108 400 1125 2646 5488 10368 18225 A019584
1 9 45 165 495 1287 3003 6435 12870 A000581
1 16 90 320 875 2016 4116 7680 13365 A119771
1 15 90 350 1050 2646 5880 11880 22275 A001297
1 12 63 224 630 1512 3234 6336 11583 A027810
1 10 55 220 715 2002 5005 11440 24310 A000582
1 24 162 640 1875 4536 9604 18432 32805 A019583
1 16 100 400 1225 3136 7056 14400 27225 A001249
1 14 84 336 1050 2772 6468 13728 27027 A027818
1 27 216 1000 3375 9261 21952 46656 91125 A059827
1 20 135 560 1750 4536 10290 21120 40095 A085284
		

Crossrefs

Cf. A064553 (second diagonal), A080688 (second diagonal resorted).

Programs

  • Maple
    A128629 := proc(n,m) if n = 1 then 1; elif isprime(n) then p := numtheory[pi](n) ; binomial(p+m-1,p) ; else a := 1 ; for p in ifactors(n)[2] do a := a* procname(op(1,p),m)^ op(2,p) ; od: fi; end: # R. J. Mathar, Sep 09 2009

Extensions

A-number added to each row of the examples by R. J. Mathar, Sep 09 2009

A135276 a(0)=0, a(1)=1; for n>1, a(n) = a(n-1) + n^0 if n odd, a(n) = a(n-1) + n^1 if n is even.

Original entry on oeis.org

0, 1, 3, 4, 8, 9, 15, 16, 24, 25, 35, 36, 48, 49, 63, 64, 80, 81, 99, 100, 120, 121, 143, 144, 168, 169, 195, 196, 224, 225, 255, 256, 288, 289, 323, 324, 360, 361, 399, 400, 440, 441, 483, 484, 528, 529, 575, 576, 624, 625, 675, 676, 728, 729, 783, 784, 840, 841, 899, 900, 960, 961
Offset: 0

Views

Author

Artur Jasinski, May 12 2008, corrected May 17 2008

Keywords

Comments

Index to family of sequences of the form a(n) = a(n-1) + n^r if n odd, a(n) = a(n-1)+ n^s if n is even, for n > 1 and a(1)=1:
s=0, s=1, s=2, s=3, s=4, s=5
r=0, A000027, this seq, A135301, A135332, A140142, A140143;
Equals triangle A070909 * [1,2,3,...]. - Gary W. Adamson, May 16 2010
Right edge of the triangle in A199332: a(n) = A199332(n,n), for n > 0. - Reinhard Zumkeller, Nov 23 2011

Crossrefs

Programs

  • Magma
    [(2*n^2+6*n+1+(2*n-1)*(-1)^n)/8 : n in [0..100]]; // Wesley Ivan Hurt, Mar 22 2016
  • Maple
    A135276:=n->( 2*n^2 + 6*n + 1 + (2*n-1)*(-1)^n )/8: seq(A135276(n), n=0..100); # Wesley Ivan Hurt, Mar 22 2016
  • Mathematica
    a = {}; r = 0; s = 1; Do[k = 0; Do[k = k + (Sin[Pi m/2]^2) m^r + (Cos[Pi m/2]^2) m^s, {m, 1, n}]; AppendTo[a, k], {n, 0, 100}]; a (* Artur Jasinski *)
    LinearRecurrence[{1, 2, -2, -1, 1}, {0, 1, 3, 4, 8}, 50] (* G. C. Greubel, Oct 08 2016 *)
  • PARI
    A135276(n)=if(n%2,((n+1)/2)^2,(n/2+1)^2-1) \\ M. F. Hasler, May 17 2008
    
  • PARI
    my(x='x+O('x^200)); concat(0, Vec(x*(1+2*x-x^2)/((1+x)^2*(1-x)^3))) \\ Altug Alkan, Mar 23 2016
    

Formula

a(n) = (n/2 + 1)^2 - 1 if n is even, ((n+1)/2)^2 if n is odd. - M. F. Hasler, May 17 2008
From R. J. Mathar, Feb 22 2009: (Start)
G.f.: x*(1+2*x-x^2)/((1+x)^2*(1-x)^3).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). (End)
a(n) = (2*n^2 + 6*n + 1 + (2*n-1)*(-1)^n)/8. - Luce ETIENNE, Jul 08 2014
a(n) = (floor(n/2)+1)^2 + (n mod 2) - 1. - Wesley Ivan Hurt, Mar 22 2016
a(n) = A004526((n+1)^2) - A004526(n+1)^2. - Bruno Berselli, Oct 21 2016
Sum_{n>=1} 1/a(n) = 3/4 + Pi^2/6. - Amiram Eldar, Sep 08 2022

Extensions

Offset corrected by R. J. Mathar, Feb 22 2009
Edited by Michel Marcus, Apr 07 2023

A164577 Integer averages of the first perfect cubes up to some n^3.

Original entry on oeis.org

1, 12, 25, 45, 112, 162, 225, 396, 507, 637, 960, 1156, 1377, 1900, 2205, 2541, 3312, 3750, 4225, 5292, 5887, 6525, 7936, 8712, 9537, 11340, 12321, 13357, 15600, 16810, 18081, 20812, 22275, 23805, 27072, 28812, 30625, 34476, 36517, 38637, 43120
Offset: 1

Views

Author

Keywords

Comments

Integers of the form A000537(k)/k, created by the k>0 listed in A042965. - R. J. Mathar, Aug 20 2009
Integers of the form (1/4)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009

Examples

			The average of the first cube is 1^3/1=1=a(1).
The average of the first two cubes is (1^3+2^3)/2=9/2, not integer, and does not contribute to the sequence.
The average of the first three cubes is (1^3+2^3+3^3)/3=12, integer, and defines a(2).
		

Crossrefs

Programs

  • Mathematica
    Timing[s=0;lst={};Do[a=(s+=n^3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n, 5!}];lst]
    With[{nn=80},Select[#[[1]]/#[[2]]&/@Thread[{Accumulate[Range[ nn]^3],Range[ nn]}],IntegerQ]] (* or *) LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{1,12,25,45,112,162,225,396,507,637},50] (* Harvey P. Dale, Mar 14 2020 *)

Formula

G.f.: ( x*(1+11*x+13*x^2+17*x^3+34*x^4+11*x^5+6*x^6+3*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011

Extensions

Changed comments to examples - R. J. Mathar, Aug 20 2009

A181134 Sum of 13th powers: a(n) = Sum_{j=0..n} j^13.

Original entry on oeis.org

0, 1, 8193, 1602516, 68711380, 1289414505, 14350108521, 111239118928, 660994932816, 3202860761145, 13202860761145, 47725572905076, 154718778284148, 457593884876401, 1251308658130545, 3197503726489920
Offset: 0

Views

Author

Bruno Berselli, Oct 05 2010 - Oct 18 2010

Keywords

Comments

This form of recurrence is a general property of the array in A103438 (sums of the first n-th powers).

Crossrefs

Cf. A010801.
Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), A023002 (m=10), A123095 (m=11), A123094 (m=12), A181134 (m=13).

Programs

  • Magma
    [(&+[j^13: j in [0..n]]): n in [0..30]]; // G. C. Greubel, Jul 21 2021
    
  • Maple
    A181134 := proc(n) (bernoulli(14,n+1) - bernoulli(14))/14 ; end proc: seq(A181134(n), n=0..10); # R. J. Mathar, Oct 14 2010
  • Mathematica
    Accumulate[Range[0,20]^13] (* Harvey P. Dale, Oct 30 2017 *)
  • Python
    A181134_list, m = [0], [6227020800, -37362124800, 97037740800, -142702560000, 130456085760, -76592355840, 28805736960, -6711344640, 901020120, -60780720, 1569750, -8190, 1, 0 , 0]
    for _ in range(10**2):
        for i in range(14):
            m[i+1]+= m[i]
        A181134_list.append(m[-1]) # Chai Wah Wu, Nov 06 2014
    
  • Sage
    [(bernoulli_polynomial(n+1, 14) - bernoulli(14))/14  for n in (0..30)] # G. C. Greubel, Jul 21 2021

Formula

For n>0, a(n) = n*A123094(n) - Sum_{i=0..n-1} A123094(i), where Sum_{i=0..n-1} A123094(i) = A253712(n-1) = (n-1)*n^2*(n+1)*(30*n^10 - 425*n^8 + 2578*n^6 - 8147*n^4 + 12874*n^2 - 7601)/5460.
a(n) = a(-n-1) = (n*(n + 1))^2*(30*n^10 + 150*n^9 + 125*n^8 - 400*n^7 - 326*n^6 + 1052*n^5 + 367*n^4 - 1786*n^3 + 202*n^2 + 1382*n - 691)/420.
G.f.: see comment of Vladeta Jovovic in A000538.
a(n) = -Sum_{j=1..13} j*Stirling1(n+1,n+1-j)*Stirling2(n+13-j,n). - Mircea Merca, Jan 25 2014

A249677 Triangle, read by rows, with row n forming the coefficients in Product_{k=0..n} (1 + k^3*x).

Original entry on oeis.org

1, 1, 1, 1, 9, 8, 1, 36, 251, 216, 1, 100, 2555, 16280, 13824, 1, 225, 15055, 335655, 2048824, 1728000, 1, 441, 63655, 3587535, 74550304, 444273984, 373248000, 1, 784, 214918, 25421200, 1305074809, 26015028256, 152759224512, 128024064000, 1, 1296, 616326, 135459216, 14320729209, 694213330464, 13472453691584, 78340747014144, 65548320768000
Offset: 0

Views

Author

Paul D. Hanna, Nov 03 2014

Keywords

Comments

Column 1 forms the squares of the triangular numbers (A000537).
Main diagonal forms the cubes of the factorial numbers (A000442).
Row sums equal Product_{k=1..n} (k^3 + 1) = n!*Product_{k=1..n} (k*(k-1) + 1) = n!*A130032(n).

Examples

			Triangle begins:
  1;
  1, 1;
  1, 9, 8;
  1, 36, 251, 216;
  1, 100, 2555, 16280, 13824;
  1, 225, 15055, 335655, 2048824, 1728000;
  1, 441, 63655, 3587535, 74550304, 444273984, 373248000;
  1, 784, 214918, 25421200, 1305074809, 26015028256, 152759224512, 128024064000;
  1, 1296, 616326, 135459216, 14320729209, 694213330464, 13472453691584, 78340747014144, 65548320768000; ...
		

Crossrefs

Programs

  • PARI
    {T(n,k)=polcoeff(prod(m=0,n,1 + m^3*x +x*O(x^n)),k)}
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
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