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.

Previous Showing 11-18 of 18 results.

A081282 Generalized centered polygonal numbers.

Original entry on oeis.org

0, 1, 6, 22, 62, 147, 308, 588, 1044, 1749, 2794, 4290, 6370, 9191, 12936, 17816, 24072, 31977, 41838, 53998, 68838, 86779, 108284, 133860, 164060, 199485, 240786, 288666, 343882, 407247, 479632, 561968, 655248, 760529, 878934, 1011654
Offset: 0

Views

Author

Paul Barry, Mar 17 2003

Keywords

Comments

Partial sums of A006007.

Crossrefs

Cf. A081284.

Programs

  • Magma
    [(n*(n+1)*(n+2)*(n^2+2*n+7))/60: n in [0..35]]; // Vincenzo Librandi, Aug 07 2013
  • Mathematica
    CoefficientList[Series[x (1 + x^2) / (1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 07 2013 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,6,22,62,147},40] (* Harvey P. Dale, Jan 09 2014 *)

Formula

a(n) = 2*C(n+3, 5) + C(n+2, 3).
G.f.: x*(1+x^2)/(1-x)^6.
a(n) = n*(n+1)*(n+2)*(n^2+2*n+7)/60. - Carl Najafi, Nov 18 2012
a(0)=0, a(1)=1, a(2)=6, a(3)=22, a(4)=62, a(5)=147, a(n)=6*a(n-1)- 15*a(n-2)+ 20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, Jan 09 2014

A081283 An interleaved sequence of pyramidal and polygonal numbers.

Original entry on oeis.org

0, 1, 1, 5, 6, 16, 20, 40, 50, 85, 105, 161, 196, 280, 336, 456, 540, 705, 825, 1045, 1210, 1496, 1716, 2080, 2366, 2821, 3185, 3745, 4200, 4880, 5440, 6256, 6936, 7905, 8721, 9861, 10830, 12160, 13300, 14840, 16170, 17941, 19481, 21505, 23276, 25576, 27600
Offset: 0

Views

Author

Paul Barry, Mar 17 2003

Keywords

Crossrefs

Programs

  • Maple
    A081283:=n->(2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(2*n^2+2*(5+(-1)^n)*n+27-11*(-1)^n)/1536: seq(A081283(n), n=0..80); # Wesley Ivan Hurt, Apr 18 2017
  • Mathematica
    CoefficientList[Series[x (1 + x^3) / ((1 - x) (1 - x^2)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 07 2013 *)

Formula

G.f.: x*(1+x^3)/((1-x)*(1-x^2)^4).
a(2*n) = A002415(n); a(2*n+1) = A006007(n+1).
a(n) = (2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(2*n^2+2*(5+(-1)^n)*n+27-11*(-1)^n)/1536. - Luce ETIENNE, Mar 11 2015

A264854 a(n) = n*(n + 1)*(11*n^2 + 11*n - 10)/24.

Original entry on oeis.org

0, 1, 14, 61, 175, 400, 791, 1414, 2346, 3675, 5500, 7931, 11089, 15106, 20125, 26300, 33796, 42789, 53466, 66025, 80675, 97636, 117139, 139426, 164750, 193375, 225576, 261639, 301861, 346550, 396025, 450616, 510664, 576521, 648550, 727125, 812631, 905464, 1006031
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 26 2015

Keywords

Comments

Partial sums of centered 11-gonal (or hendecagonal) pyramidal numbers.

Crossrefs

Cf. A004467.
Cf. similar sequences provided by the partial sums of centered k-gonal pyramidal numbers: A006522 (k=1), A006007 (k=2), A002817 (k=3), A006325 (k=4), A006322 (k=5), A000537 (k=6), A006323 (k=7), A006324 (k=8), A236770 (k=9), A264853 (k=10), this sequence (k=11), A062392 (k=12), A264888 (k=13).

Programs

  • Magma
    [n*(n+1)*(11*n^2+11*n-10)/24: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015
    
  • Mathematica
    Table[n (n + 1) (11 n^2 + 11 n - 10)/24, {n, 0, 50}]
  • PARI
    a(n)=n*(n+1)*(11*n^2+11*n-10)/24 \\ Charles R Greathouse IV, Jul 26 2016

Formula

G.f.: x*(1 + 9*x + x^2)/(1 - x)^5.
a(n) = Sum_{k = 0..n} A004467(k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015

A129687 A129686 * A007318.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 4, 3, 1, 2, 6, 7, 4, 1, 2, 8, 13, 11, 5, 1, 2, 10, 21, 24, 16, 6, 1, 2, 12, 31, 45, 40, 22, 7, 1, 2, 14, 43, 76, 85, 62, 29, 8, 1, 2, 16, 57, 119, 161, 147, 91, 37, 9, 1, 2, 18, 73, 176, 280, 308, 238, 128, 46, 10, 1, 2, 20, 91, 249, 456
Offset: 0

Views

Author

Gary W. Adamson, Apr 28 2007

Keywords

Comments

Row sums = A084215: (1, 2, 5, 10, 20, 40, 80, ...). A007318 * A129686 = A124725.
From Philippe Deléham, Feb 12 2014: (Start)
Riordan array ((1+x^2)/(1-x), x/(1-x)).
Diagonal sums are A000032(n) - 0^n (cf. A000204).
T(n,0) = A046698(n+1).
T(n+1,1) = A004277(n).
T(n+2,2) = A002061(n+1).
T(n+3,3) = A006527(n+1) = A167875(n).
T(n+4,4) = A006007(n+1).
T(n+5,5) = A081282(n+1). (End)

Examples

			First few rows of the triangle:
  1;
  1,   1;
  2,   2,   1;
  2,   4,   3,   1;
  2,   6,   7,   4,   1;
  2,   8,  13,  11,   5,   1;
  2,  10,  21,  24,  16,   6,   1;
  2,  12,  31,  45,  40,  22,   7,   1;
  2,  14,  43,  76,  85,  62,  29,   8,   1;
  2,  16,  57, 119, 161, 147,  91,  37,   9,   1;
  ...
		

Crossrefs

Formula

A129686 * A007318 (Pascal's Triangle), as infinite lower triangular matrices.
T(n,k) = T(n-1,k) + T(n-1,k-1), T(0,0) = T(1,0) = T(1,1) = T(2,2) = 1, T(2,0) = T(2,1) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Feb 12 2014

Extensions

More terms from Philippe Deléham, Feb 12 2014

A153977 One-fourth of partial sums of A153976.

Original entry on oeis.org

2, 9, 27, 65, 135, 252, 434, 702, 1080, 1595, 2277, 3159, 4277, 5670, 7380, 9452, 11934, 14877, 18335, 22365, 27027, 32384, 38502, 45450, 53300, 62127, 72009, 83027, 95265, 108810, 123752, 140184, 158202, 177905, 199395, 222777, 248159, 275652, 305370, 337430
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Maple
    A153977:=n->(1/4)*sum(i^3 + (i+2)^3, i=0..n): seq(A153977(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017
  • Mathematica
    a[n_]:=n^3;lst={};s=0;Do[s+=(a[n]+a[n+2]);AppendTo[lst,s/4],{n,0,6!}];lst
    Accumulate[Array[#^3+(#+2)^3&,40,0]]/4 (* or *) LinearRecurrence[ {5,-10,10,-5,1},{2,9,27,65,135},40] (* Harvey P. Dale, Aug 02 2011 *)
  • PARI
    a(n)=(n^4 + 2*n^3 + 7*n^2 + 6*n)/8 \\ Charles R Greathouse IV, Feb 06 2017

Formula

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5, a(1)=2, a(2)=9, a(3)=27, a(4)=65, a(5)=135. - Harvey P. Dale, Aug 02 2011
a(n) = (A000217(n-1)^2 + A000217(n+1)^2 - 1)/4. - Richard R. Forberg, Dec 25 2013
Recurrence: (n-1)*(n^2 - n + 6)*a(n) = (n+1)*(n^2 + n + 6)*a(n-1). - Vaclav Kotesovec, Dec 26 2013
a(n) = A000217(A000217(n)) + A000217(n). - Bruno Berselli, May 28 2015
a(n) = (A000217(n)^2 + 3*A000217(n))/2 where A000217(n) is the n-th triangular number. - Frederic Isenmann, Feb 04 2017
Sum_{n>=1} 1/a(n) = 14/9 - 4*tanh(sqrt(23)*Pi/2)*Pi/(3*sqrt(23)). - Amiram Eldar, Aug 23 2022
From Elmo R. Oliveira, Aug 28 2025: (Start)
G.f.: x*(2 - x + 2*x^2)/(1-x)^5.
E.g.f.: x*(2 + x)^2*(4 + x)*exp(x)/8. (End)

A067956 Number of nodes in virtual, "optimal", chordal graphs of diameter 4 and degree n+1.

Original entry on oeis.org

9, 16, 41, 66, 129, 192, 321, 450, 681, 912, 1289, 1666, 2241, 2816, 3649, 4482, 5641, 6800, 8361, 9922, 11969, 14016, 16641, 19266, 22569, 25872, 29961, 34050, 39041, 44032, 50049, 56066, 63241, 70416, 78889
Offset: 1

Views

Author

S. Bujnowski & B. Dubalski (slawb(AT)atr.bydgoszcz.pl), Mar 08 2002

Keywords

Examples

			For n=5, n is odd; t=3; a(5) = (2*3*(3+1)*(3^2+3+4)/3)+1 = ((6*4*16)/3)+1 = 129.
For n=6, n is even; t=3; a(6) = a(5) + ((2*3+1)*(2*t^2+2*t+3))/3 = 129 + (7*27)/3 = 192.
		

References

  • Concrete Mathematics, R. L. Graham, D. E. Knuth, O. Patashnik, 1994, Addison-Wesley Company, Inc.

Crossrefs

Cf. A006007.

Programs

  • Maple
    for n from 1 to k do if ((n mod 2 ) = 1) then t := (n+1)/2; a[n] := ((2*(t*(t+1)*(t^2+t+4))/3)+1); else t := (n)/2; a[n] := ((2*(t*(t+1)*(t^2+t+4)/3)+1)+(2*t+1)*(2*t^2+2*t+3)/3); fi; print(a[n]); od;
  • Mathematica
    Array[((2 #2 (#2 + 1) (#2^2 + #2 + 4))/3) + 1 + (Boole[EvenQ[#1]]*((2 #2 + 1) (2 #2^2 + 2 #2 + 3))/3) & @@ {#, (# + Boole[OddQ[#]])/2} &, 35] (* Michael De Vlieger, Jul 29 2022 *)

Formula

For n odd, t = (n+1)/2, a(n) = ((2*t*(t+1)*(t^2+t+4))/3)+1;
for n even, t = n/2, a(n) = (((2*t*(t+1)*(t^2+t+4))/3)+1)+((2*t+1)*(2*t^2+2*t+3))/3.
G.f.: -x*(9-2*x-9*x^2+6*x^3+11*x^4-6*x^5-3*x^6+2*x^7)/(1+x)^3/(x-1)^5 . - R. J. Mathar, Apr 07 2025

A355010 Array read by ascending antidiagonals: T(n, k) is the number of n-core partitions with k corners.

Original entry on oeis.org

1, 3, 1, 6, 5, 1, 10, 16, 7, 1, 15, 40, 31, 9, 1, 21, 85, 105, 51, 11, 1, 28, 161, 295, 219, 76, 13, 1, 36, 280, 721, 771, 396, 106, 15, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1, 55, 705, 3186, 6244, 6083, 3235, 995, 181, 19, 1, 66, 1045, 5985, 15156, 19348, 13663, 5685, 1445, 226, 21, 1
Offset: 2

Views

Author

Stefano Spezia, Jun 15 2022

Keywords

Comments

T(n, k) is also equal to the number of cornerless Motzkin paths of length 2*k + n - 1 with n - 1 flat steps (see Theorem 3.3 and Proposition 3.4 at pp. 13 - 14 in Cho et al.).
In proposition 3.4 in Cho et al., the Narayana number is defined as N(k, i) = binomial(k, i)*binomial(k, i-1)/k, unlike A001263.

Examples

			The array begins:
    1,  1,   1,   1,    1,    1,    1,    1, ...
    3,  5,   7,   9,   11,   13,   15,   17, ...
    6, 16,  31,  51,   76,  106,  141,  181, ...
   10, 40, 105, 219,  396,  650,  995, 1445, ...
   15, 85, 295, 771, 1681, 3235, 5685, 9325, ...
   ...
		

Crossrefs

Cf. A000012 (n = 2), A001263, A005408 (n = 3), A005891 (n = 4), A006007, A063490 (n = 5), A160747 (n = 6), A161680 (k = 1), A355011.

Programs

  • Mathematica
    T[n_,k_]:=Sum[Binomial[k,i]Binomial[k,i-1]Binomial[n+2(k-i),2k]/k,{i,Min[k,Floor[n/2]]}]; Flatten[Table[T[n-k+1,k],{n,2,12},{k,1,n-1}]]

Formula

T(n, k) = Sum_{i=1..min(k,floor(n/2))} N(k, i)*binomial(n+2*(k-i), 2*k), where N(k, i) = binomial(k, i)*binomial(k, i-1)/k. (See proposition 3.4 in Cho et al.).
T(n, 2) = A006007(n-1).

A383957 Sum of the legs of the unique primitive Pythagorean triple whose inradius is A000108(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

7, 7, 17, 71, 449, 3697, 35377, 369799, 4095521, 47297537, 564278417, 6911822737, 86538816337, 1103803791601, 14305269324961, 187980077927431, 2500329797088481, 33615543666867361, 456277457385934801, 6246438372527004961, 86175353802778434481, 1197196443885744428881, 16738118900659230353761
Offset: 0

Views

Author

Keywords

Examples

			For n=1, the short leg is A383251(1,1) = 3 and the long leg is A383251(1,2) = 4 so the sum of the legs is then a(1) = 3 + 4 = 7.
		

Crossrefs

Programs

  • Mathematica
    a=Table[(2n)!/(n!(n+1)!),{n,0,23}];Apply[Join,Map[{2#^2+4#+1}&,a]]

Formula

a(n) = A383251(n,1) + A383251(n,2).
a(n) = 2*(A000108(n))^2 + 4*A000108(n) + 1.
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