A248851 -id:A248851 - cp's OEIS frontend

A117143 Number of partitions of n in which any two parts differ by at most 3.

1, 2, 3, 5, 7, 10, 13, 17, 22, 27, 33, 41, 48, 57, 68, 78, 90, 105, 118, 134, 153, 170, 190, 214, 235, 260, 289, 315, 345, 380, 411, 447, 488, 525, 567, 615, 658, 707, 762, 812, 868, 931, 988, 1052, 1123, 1188, 1260, 1340, 1413, 1494, 1583, 1665, 1755, 1854

Offset: 1

Emeric Deutsch, Feb 27 2006

			a(6) = 10 because we have [6], [4,2], [4,1,1], [3,3], [3,2,1], [3,1,1,1], [2,2,2], [2,2,1,1], [2,1,1,1,1] and [1,1,1,1,1,1] ([5,1] does not qualify).

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Jonathan Bloom, Nathan McNew, Counting pattern-avoiding integer partitions, arXiv:1908.03953 [math.CO], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,-2,-2,1,1,1,-1).

Cf. A117142.
Column k=3 of A194621.
Cf. A002717, A045947, A248851.

[(2*Floor((n+2)/3)*(14*Floor((n+2)/3)^2-(10*n+21)*Floor((n+2)/3)+2*(n^2+5*n+7))-(1-(-1)^Floor((n+2)/3))*(-1)^(n+2-Floor((n+2)/3)))/16: n in [1..60]]; // Vincenzo Librandi, May 12 2015

g:=sum(x^k/(1-x^k)/(1-x^(k+1))/(1-x^(k+2))/(1-x^(k+3)),k=1..85): gser:=series(g,x=0,65): seq(coeff(gser,x^n),n=1..59); with(combinat): for n from 1 to 7 do P:=partition(n): A:={}: for j from 1 to nops(P) do if P[j][nops(P[j])]-P[j][1]<4 then A:=A union {P[j]} else A:=A fi od: print(A); od: # this program yields the partitions

Table[Count[IntegerPartitions[n], ?(Max[#] - Min[#] <= 3 &)], {n, 30}] (* _Birkas Gyorgy, Feb 20 2011 *)

Vec(x*(x^5-x^4-x^3+x+1)/((x-1)^4*(x+1)*(x^2+x+1)^2) + O(x^100)) \\ Colin Barker, Mar 05 2015

G.f.: sum(x^k/[(1-x^k)(1-x^(k+1))(1-x^(k+2))(1-x^(k+3))], k=1..infinity). More generally, the g.f. of the number of partitions in which any two parts differ by at most b is sum(x^k/product(1-x^j, j=k..k+b), k=1..infinity).
G.f.: x*(x^5-x^4-x^3+x+1) / ((x-1)^4*(x+1)*(x^2+x+1)^2). - Colin Barker, Mar 05 2015
a(n)=(2*floor((n+2)/3)*(14*floor((n+2)/3)^2-(10*n+21)*floor((n+2)/3)+2*(n^2+5*n+7))-(1-(-1)^floor((n+2)/3))*(-1)^(n+2-floor((n+2)/3)))/16. - Luce ETIENNE, May 12 2015

A256666 a(n) = ( 2n(2n^2 + 11n + 26) - (-1)^n + 1 )/16.

Original entry on oeis.org

0, 5, 14, 29, 51, 82, 123, 176, 242, 323, 420, 535, 669, 824, 1001, 1202, 1428, 1681, 1962, 2273, 2615, 2990, 3399, 3844, 4326, 4847, 5408, 6011, 6657, 7348, 8085, 8870, 9704, 10589, 11526, 12517, 13563, 14666, 15827, 17048, 18330, 19675, 21084, 22559

Offset: 0

Luce ETIENNE, Apr 07 2015

Consider a grid of small triangles of side 1 forming polygon with side n*(n+3): a(n) is the number of equilateral triangles of side length >=1 in this figure that are oriented with the sides of figure.
This sequence gives the number of triangles of all sizes in a ((n^2+3*n))-iamond with a 3*(2*n-1)-gon n>=1.
Equals (1/2)*Sum_{i=0..n-1} (n-i)*(n+1-i) + (-3 + (1/8)*Sum_{j=0..(2*n+3+(-1)^n)/4} (2*n+5-(-1)^n-4*j)*(2 n+5+(-1)^n-4*j) ) numbers of triangles in a direction and in the opposite direction.
It is also a way (3 stages) to surround triangular n^2-iamonds by 3*n triangles side 1: in first stage we obtain A045947, in second stage A248851, in third stage this sequence.

			From third comment: a(0)=0, a(1)=1+4, a(2)=4+10, a(3)=10+19, a(4)=20+31, a(5)=35+47, a(6)=56+67.

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A002623, A028552, A045947, A117143, A164004, A248851.

[(4*n^3+22*n^2+52*n+1-(-1)^n)/16: n in [0..50]]; // Vincenzo Librandi, Apr 08 2015

Table[(4 n^3 + 22 n^2 + 52 n + 1 - (-1)^n)/16, {n, 0, 50}] (* Vincenzo Librandi, Apr 08 2015 *)
LinearRecurrence[{3,-2,-2,3,-1},{0,5,14,29,51},50] (* Harvey P. Dale, Aug 18 2020 *)

concat(0, Vec(x*(2*x^3-3*x^2-x+5)/((x-1)^4*(x+1)) + O(x^100))) \\ Colin Barker, Apr 07 2015

a(n) = 2*A248851(n) - A045947(n) + A004526(n+1).
G.f.: x*(2*x^3-3*x^2-x+5) / ((x-1)^4*(x+1)). - Colin Barker, Apr 07 2015
a(n) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5) for n>4. - Colin Barker, Apr 07 2015

A265282 Number of triangles in a certain geometric structure: see "Illustration of initial terms" link for precise definition.

Original entry on oeis.org

0, 1, 3, 5, 10, 13, 22, 26, 41, 46, 68, 74, 105, 112, 153, 161, 214, 223, 289, 299, 380, 391, 488, 500, 615, 628, 762, 776, 931, 946, 1123, 1139, 1340, 1357, 1583, 1601, 1854, 1873, 2154, 2174, 2485, 2506, 2848, 2870, 3245, 3268, 3677, 3701, 4146, 4171, 4653

Offset: 0

Luce ETIENNE, Dec 06 2015

In words: This sequence gives the number of triangles of all sizes in a (2*n^2+8*n-1+(-1)^n)/8-polyiamond with a (7*n-2-(n-2)*(-1)^n)/4-gon: we have (2*n^3+9*n^2+31*n+21+3*(n^2-5*n-7)*(-1)^n)/96 triangles in a direction and (2*n^3+27*n^2+109*n-66+3*(n^2+9*n+18)*(-1)^n+12*(-1)^((2*n-1+(-1)^n)/4))/192 triangles in the other direction. (But the Illustration link is far more informative. - N. J. A. Sloane, Jan 23 2016)
At stage n, we count (2*n^2 + 6*n + 3 - (2*n+3)*(-1)^n)/16 triangles of size 1 in one direction and (2*n^2 + 10*n - 5 + (2*n+5)*(-1)^n)/16 triangles of size 1 in the opposite direction. The total number of triangles of size 1 in both directions is A024206(n+1).
We observe that a(4)=10 strengthens the Pythagorean relation between 4 and 10 (Tetraktys): cf. triangular numbers, A000217; and that it is from n = 4 we can see and count hexagonal and dodecagonal forms, for example, in a reticular system (incomplete with hexagonal holes) by opposition to the compact shape obtained from A002717.
We can obtain this reticular system from A248851.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Luce ETIENNE, Illustration of initial terms
Luce ETIENNE, A265282 from A248851
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,0,0,-2,2,1,-1).

Cf. A000217, A000292, A001477, A001859, A002620, A002717, A004006, A004526, A045947, A132337.

Cf. A248851.

[(2*n^3 + 15*n^2 + 57*n - 8 + (3*n^2 - n + 4)*(-1)^n + 4*(-1)^((2*n - 1 + (-1)^n) div 4)) / 64: n in [0..50]]; // Vincenzo Librandi, Dec 07 2015

Table[(2*n^3 + 15*n^2 + 57*n - 8 + (3*n^2 - n + 4)*(-1)^n +
    4*(-1)^((2*n - 1 + (-1)^n)/4))/64, {n, 0, 100}] (* G. C. Greubel, Dec 20 2015 *)
LinearRecurrence[{1,2,-2,0,0,-2,2,1,-1},{0,1,3,5,10,13,22,26,41},60] (* Harvey P. Dale, Aug 07 2019 *)

vector(100, n, n--; (2*n^3+15*n^2+57*n-8+(3*n^2-n+4)*(-1)^n+4*(-1)^((2*n-1+(-1)^n)/4))/64) \\ Altug Alkan, Dec 06 2015

concat(0, Vec(x*(1+2*x+x^3-x^4-x^5+x^7)/((1-x)^4*(1+x)^3*(1+x^2)) + O(x^100))) \\ Colin Barker, Dec 07 2015

a(n) = A045947(floor(n/2)) + A024206(n+1). Note that A045947(floor(n/2)) = (2*n^3-n^2-7*n+(3*n^2-n-4)*(-1)^n+4*(-1)^((2*n-1+(-1)^n)/4))/64.
a(n) = (2*n^3 + 15*n^2 + 57*n - 8 + (3*n^2 - n + 4)*(-1)^n + 4*(-1)^((2*n - 1 + (-1)^n)/4))/64.
G.f.: x*(1+2*x+x^3-x^4-x^5+x^7) / ((1-x)^4*(1+x)^3*(1+x^2)). - Colin Barker, Dec 07 2015

a(26) corrected by Altug Alkan, Dec 06 2015

cp's OEIS Frontend

A117143 Number of partitions of n in which any two parts differ by at most 3.

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A256666 a(n) = ( 2*n*(2*n^2 + 11*n + 26) - (-1)^n + 1 )/16.

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A265282 Number of triangles in a certain geometric structure: see "Illustration of initial terms" link for precise definition.

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Magma

Mathematica

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PARI

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A256666 a(n) = ( 2n(2n^2 + 11n + 26) - (-1)^n + 1 )/16.