A049450 -id:A049450 - cp's OEIS frontend

A272134 a(n) = n(15n^2 - 15*n + 4).

0, 4, 68, 282, 736, 1520, 2724, 4438, 6752, 9756, 13540, 18194, 23808, 30472, 38276, 47310, 57664, 69428, 82692, 97546, 114080, 132384, 152548, 174662, 198816, 225100, 253604, 284418, 317632, 353336, 391620, 432574, 476288, 522852, 572356, 624890, 680544

Offset: 0

Vincenzo Librandi, Apr 27 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16)
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A049450, A272357.

Magma
```
[n*(15*n^2-15*n+4): n in [0..40]];
```

Table[n (15 n^2 - 15 n + 4), {n, 0, 40}]

vector(100, n, n--; n*(15*n^2 - 15*n + 4)) \\ Altug Alkan, Apr 28 2016

for n in range(0,10**3):print(n*(15*n**2-15*n+4),end=", ") # Soumil Mandal, Apr 30 2016

O.g.f.: 2*x*(2 + 26*x + 17*x^2)/(1-x)^4.
E.g.f.: x*(4 + 30*x + 15*x^2)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3.
See page 7 in Brent's paper:
a(n) = 2*n^2*A049450(n) - n*(2*n-1)*A049450(n-1).
A272357(n) = 2*n^2*a(n) - n*(2*n-1)*a(n-1).

A294513 Denominators of the partial sums of the reciprocals of twice the pentagonal numbers.

Original entry on oeis.org

2, 5, 120, 1320, 9240, 52360, 52360, 602140, 70450380, 2043061020, 16344488160, 3268897632, 62109055008, 2546471255328, 1157486934240, 54401885909280, 272009429546400, 4805499921986400, 4805499921986400, 283524495397197600, 418536159872053600

Offset: 0

Wolfdieter Lang, Nov 02 2017

The corresponding numerators are given by A250328(n+1), n >= 0.
Twice the positive pentagonal numbers are A049450(k+1) = (k+1)*(3*k+2), k >= 0.
For the general case V(m,r;n) = Sum_{k=0..n} 1/((k + 1)*(m*k + r)) = (1/(m - r))*Sum_{k=0..n} (m/(m*k + r) - 1/(k+1)), for r = 1, ..., m-1 and m = 2, 3, ..., and their limits see a comment in A294512. Here [m,r] = [3,2].
The limit of the series is V(3,2) = lim_{n -> oo} V(3,2;n) = (3/2)*log(3) - Pi/(2*sqrt(3)) = 0.74101875088505561179... given in A294514.

			The rationals V(3,2;n), n >= 0, begin: 1/2, 3/5, 77/120, 877/1320, 6271/9240, 36049/52360, 36423/52360, 422137/602140, 49691099/70450380, 1448086909/2043061020, ...
V(3,2;10^4) = 0.7409854223(Maple, 10 digits) to be compared with 0.7410187513 from V(3,2) given in A294514.
Conjecture tests: a(0) = 2 =  A250327(1)/1, 2* a(1) = 5 = 2*A250327(2)/2 = A250327(2), a(2) = 120 = 2*A250327(2)/3 = 2*180/3, ...

Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, pp. 189 - 193 (with v_m(r) = ((m-r)/m)*V(m,r)).

Cf. A049450, A250327(n+1), A250328(n+1), A294512.

Denominator@ Accumulate@ Array[1/(2 PolygonalNumber[5, #]) &, 21] (* Michael De Vlieger, Nov 02 2017 *)

a(n) = denominator(V(3,2;n)) with V(3,2;n) = Sum_{k=0..n} 1/((k + 1)*(3*k + 2)) = Sum_{k=0..n} 1/A049450(k+1) = Sum_{k=0..n} (3/(3*k + 2) - 1/(k+1)).

a(n) = 2*A250327(n+1)/(n+1) [conjecture].

A306249 Number of ways to write n as x(2x-1) + y(3y-1) + z(4z-1) + w(5w-1), where x,y,z are nonnegative integers and w is 0 or 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 1, 2, 3, 4, 2, 3, 3, 3, 4, 2, 2, 1, 4, 3, 1, 1, 5, 4, 3, 3, 3, 4, 4, 3, 1, 3, 3, 5, 1, 2, 4, 5, 4, 4, 2, 3, 7, 3, 3, 2, 5, 3, 3, 2, 2, 3, 4, 5, 1, 4, 6, 6, 2, 3, 5, 3, 3, 3, 5, 4, 5, 5, 3, 6, 6, 4, 3, 4, 5, 2, 3, 4, 4, 5, 2, 2, 5, 6, 6, 1, 5, 3, 6, 2, 4, 3, 4, 4, 2

Offset: 0

Zhi-Wei Sun, Jan 31 2019

nonn

Conjecture: a(n) > 0 for any nonnegative integer n.
This has been verified for n up to 10^6. By Theorem 1.2 of the linked 2017 paper of the author, any nonnegative integer can be written as x*(2x-1) + y*(3y-1) + z*(4z-1) with x,y,z integers.
We have some other similar conjectures. For example, we conjecture that each n = 0,1,2,... can be written as x*(3x-1)/2 + y*(5y-1)/2 + z*(7z-1)/2 + w*(9w-1)/2) (or x*(x-1) + y*(2y-1) + z*(3z-1) + w*(4w-1)) with x,y,z,w nonnegative integers.

			a(1) = 1 with 1 = 1*(2*1-1) + 0*(3*0-1) + 0*(4*0-1) + 0*(5*0-1).
a(2) = 1 with 2 = 0*(2*0-1) + 1*(3*1-1) + 0*(4*0-1) + 0*(5*0-1).
a(12) = 1 with 12 = 2*(2*2-1) + 1*(3*1-1) + 0*(4*0-1) + 1*(5*1-1).
a(26) = 1 with 26 = 2*(2*2-1) + 1*(3*1-1) + 2*(4*2-1) + 1*(5*1-1).
a(220) = 1 with 220 = 6*(2*6-1) + 7*(3*7-1) + 2*(4*2-1) + 0*(5*0-1).
a(561) = 1 with 561 = 17*(2*17-1) + 0*(3*0-1) + 0*(4*0-1) + 0*(5*0-1).
a(1356) = 1 with 1356 = 23*(2*23-1) + 1*(3*1-1) + 9*(4*9-1) + 1*(5*1-1).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Cf. A000326, A000384, A002378, A033991, A049450, A255350, A306225, A306227, A306239, A306242.

HexQ[n_]:=HexQ[n]=IntegerQ[Sqrt[8n+1]]&&(n==0||Mod[Sqrt[8n+1]+1,4]==0);
tab={};Do[r=0;Do[If[HexQ[n-x(5x-1)-y(4y-1)-z(3z-1)],r=r+1],{x,0,Min[1,(Sqrt[20n+1]+1)/10]},{y,0,(Sqrt[16(n-x(5x-1))+1]+1)/8},{z,0,(Sqrt[12(n-x(5x-1)-y(4y-1))+1]+1)/6}];tab=Append[tab,r],{n,0,100}];Print[tab]

A306250 Number of ways to write n as x(3x+1) + y(3y-1) + z(3z+2) + w(3w-2), where x,y,z,w are nonnegative integers with xyz = 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 3, 1, 1, 1, 3, 4, 3, 3, 2, 2, 2, 2, 2, 2, 4, 3, 3, 3, 3, 2, 4, 3, 3, 2, 2, 4, 4, 4, 4, 2, 5, 4, 1, 3, 3, 5, 3, 4, 4, 4, 3, 3, 2, 2, 6, 4, 6, 4, 6, 4, 4, 4, 3, 2, 5, 4, 4, 3, 5, 4, 7, 4, 2, 2, 4, 8, 3, 4, 6, 4, 5, 6, 3, 5, 5, 6, 6, 5, 4, 5, 3, 4, 2, 4, 5, 6, 6, 7, 6, 1, 8

Offset: 0

Zhi-Wei Sun, Feb 01 2019

nonn

Conjecture: a(n) > 0 for any nonnegative integer n.

Clearly, a(n) <= A306242(n). We have verified a(n) > 0 for all n = 0..10^6.

			a(12) = 1 with 12 = 1*(3*1+1) + 0*(3*0-1) + 0*(3*0+2) + 2*(3*2-2).
a(42) = 1 with 42 = 0*(3*0+1) + 1*(3*1-1) + 0*(3*0+2) + 4*(3*4-2).
a(62) = 3 with 62 = 3*(3*3+1) + 3*(3*3-1) + 0*(3*0+2) + 2*(3*2-2)
= 4*(3*4+1) + 2*(3*2-1) + 0*(3*0+2) + 0*(3*0-2) = 4*(3*4+1) + 1*(3*1-1) + 0*(3*0+2) + 2*(3*2-2).
a(99) = 1 with 99 = 2*(3*2+1) + 0*(3*0-1) + 5*(3*5+2) + 0*(3*0-2).
a(118) = 1 with 118 = 0*(3*0+1) + 6*(3*6-1) + 2*(3*2+2) + 0*(3*0-2).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Cf. A000567, A045944, A049450, A049451, A255350, A306242.

OctQ[n_]:=OctQ[n]=IntegerQ[Sqrt[3n+1]]&&(n==0||Mod[Sqrt[3n+1]+1,3]==0);
tab={};Do[r=0;Do[If[OctQ[n-x(3x+2)-y(3y+1)-z(3z-1)],r=r+1],{x,0,(Sqrt[3n+1]-1)/3},{y,0,(Sqrt[12(n-x(3x+2))+1]-1)/6},{z,0,If[x>0&&y>0,0,(Sqrt[12(n-x(3x+2)-y(3y+1))+1]+1)/6]}];tab=Append[tab,r],{n,0,100}];Print[tab]

A069741 Let M_n be the n X n matrix M_(i,j)=1/(2^i+2^j), then a(n) is the numerator of det(M_n).

Original entry on oeis.org

1, 1, 1, 49, 2401, 113060689, 260871824431729, 9708455965188246321478801, 361304320362377236050632364626862769, 3511057522394397982450601057907077808699210592028881

Offset: 1

Benoit Cloitre, Apr 21 2002

a(n) seems always to be a square and 7 seems to follow a rule in a(n) factorization. Maximal k such that 7^k divides a(n) are 0, 0, 0, 2, 4, 6, 10, 14, 18, 24, 30, 36, 44, 52, 60, 70, 80, 90, 102, 114, 126, 142, 158, 174, 192... Hence if b(n)=maximum exponent of 7 in factorization of a(n), b(3n+1)=A049450(n); b(3n+2)=A049450(n)+2*n; b(3n+3)=A049450(n)+4n

Vincenzo Librandi, Table of n, a(n) for n = 1..25

Cf. A069743.

for(n=1,70,print1(numerator(matdet(matrix(n,n,i,j,1/(2^i+2^j)))),","))

A165410 Hankel transform of the transform of 2^n given by A165409.

Original entry on oeis.org

1, 0, -4, -16, 0, 1024, 16384, 0, -16777216, -1073741824, 0, 17592186044416, 4503599627370496, 0, -1180591620717411303424, -1208925819614629174706176, 0, 5070602400912917605986812821504, 20769187434139310514121985316880384

Offset: 0

Paul Barry, Sep 17 2009

sign

Powers of two occurring in this sequence are based on the hexagonal spiral pattern of A049450 and A049451 (see A152749):
.
        16--15--14
        /         \
      17   5---4  13
      /   /     \   \
    18   6   0   3  12
    /   /   /   /   /
  19   7   1---2  11  26
    \   \         /   /
    20   8---9--10  25
      \             /
      21--22--23--24
.
The powers, (0,-oo,2,4,-oo,10,14,-oo,24,30,-oo,...) correspond to vertically joining pairs on the (0,4) and (0,2) radial lines, with -oo corresponding to the jump to the next pair.
The Hankel transforms of transforms of r^n behave similarly -- we get 1, 0, -r^2, -r^4, 0, r^10, r^14, ....
Note the Somos-4 property: a(3n) = 4*a(3n-1)*a(3n-3)/e(3n-4). Related to elliptic curve y^2 = 1 - 8x^3 in g.f. of A165409.

Cf. A165409.

A290168 If n is even then a(n) = n^2(n+2)/8, otherwise a(n) = (n-1)n*(n+1)/8.

Original entry on oeis.org

0, 0, 2, 3, 12, 15, 36, 42, 80, 90, 150, 165, 252, 273, 392, 420, 576, 612, 810, 855, 1100, 1155, 1452, 1518, 1872, 1950, 2366, 2457, 2940, 3045, 3600, 3720, 4352, 4488, 5202, 5355, 6156, 6327, 7220, 7410, 8400

Offset: 0

Jean-François Alcover and Paul Curtz, Jul 23 2017

nonn

Bisection of a(n) [0, 2, 12, 36, 80, 150, 252, ...] is A011379.
Bisection [0, 3, 15, 42, 90, 165, 273, ...] is A059270.
Considering s(n) = [0, 0, 0, 0, 1, 1, 3, 3, 6, 6, 10, 10, 15, 15, ...] (triangular numbers repeated - see A008805), a(n) = n*s(n+2) holds.
Considering the first differences of a(n), b(n) = [0, 2, 1 , 9, 3, 21, 6, 38, 10, 60, 15, 87, ...], b(n) shows bisections A000217 and A005476. In addition, b(n) begins like A249264 up to 12th term, and is an alternation of 4 multiples of 3 and 2 not multiples; b(n) is also such that b(2n) + b(2n+1) = A049450(n).
Considering the second differences c(n), c(n) shows bisections A001105(n+1) and -A000384(n+1), c(n) has 3 consecutive terms multiples of 3 alternating with 3 not multiples; in addition, c(2n) + c(2n+1) = A000027(n).
Considering a(n)/c(n) = [0, 0, 1/4, -1/2, 2/3, -1, 9/8, -3/2, 8/5, -2, 25/12, -5/2, ...], it appears that it is A129194(n)/A022998(n+1) and -A026741(n)/A000034(n) alternating.

Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A000027, A000034, A000217, A000384, A001105, A005476, A008805, A011379, A022998, A026741, A049450, A059270, A129194, A135713, A161680, A249264.

a[n_] := If[EvenQ[n], n^2*(n + 2)/8, (n - 1)*n*(n + 1)/8]; Table[a[n], {n, 0, 40}]

a(n) = if(n%2==0, n^2*(n+2)/8, (n-1)*n*(n+1)/8) \\ Felix Fröhlich, Jul 23 2017

G.f.: x^2*(2 + x + 3*x^2)/((x-1)^4*(x+1)^3).
a(n) = (1/16)*(-1)^n*n*(1 + (-1)^(n+1) + 2*(1 + (-1)^n)*n + 2*(-1)^n*n^2).
Sum_{n>=2} 1/a(n) = 5 + Pi^2/6 - 8*log(2). - Amiram Eldar, Sep 17 2022

A256560 Triangle read by rows, sums of 2 distinct nonzero squares plus sums of 2 distinct nonzero cubes: T(n,k) = n^2 + k^2 + n^3 + k^3, 1 <= k <= n-1.

Original entry on oeis.org

14, 38, 48, 82, 92, 116, 152, 162, 186, 230, 254, 264, 288, 332, 402, 394, 404, 428, 472, 542, 644, 578, 588, 612, 656, 726, 828, 968, 812, 822, 846, 890, 960, 1062, 1202, 1386, 1102, 1112, 1136, 1180, 1250, 1352, 1492, 1676, 1910

Offset: 2

Bob Selcoe, Apr 02 2015

All terms are even.
T(n,1) = A011379(n) + 2.
When n=k+1, T(n,k+1) = A011379(n-1) + A011379(n) = 2n^3 - n^2 + n.

			Triangle starts T(2,1):
n\k   1    2    3    4    5    6    7     8    9   10
2:   14
3:   38   48
4:   82   92   116
5:   152  162  186  230
6:   254  264  288  332  402
7:   394  404  428  472  542  644
8:   578  588  612  656  726  828  968
9:   812  822  846  890  960  1062 1202 1386
10:  1102 1112 1136 1180 1250 1352 1492 1676 1910
11:  1454 1464 1488 1532 1602 1704 1844 2028 2262 2552
...
The successive terms are: (2^2 + 1^2 + 2^3 + 1^3), (3^2 + 1^2 + 3^3 + 1^3), (3^2 + 2^2 + 3^3 + 2^3), (4^2 + 1^2 + 4^3 + 1^3), (4^2 + 2^2 + 4^3 + 2^3), (4^2 + 3^2 + 4^3 + 3^3), ...
T(7,4) = 472 because 7^2 + 7^3 + 4^2 + 4^3 = 472.

Cf. A055096 (sums of 2 distinct nonzero squares), A256497 (sums of 2 distinct nonzero cubes), A011379, A024670, A004431, A049450.

a(n) = A055096(n) + A256497(n-1).
T(n,k) = T055096(n,k) + T256547(n-1,k).
T(n,k) = T(n-1,k) + A049450(n).
T(n,k) = T(n,k-1) + A049450(k).
T(n,k) = A011379(n) + A011379(k).

cp's OEIS Frontend

A272134 a(n) = n*(15*n^2 - 15*n + 4).

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A294513 Denominators of the partial sums of the reciprocals of twice the pentagonal numbers.

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A306249 Number of ways to write n as x*(2x-1) + y*(3y-1) + z*(4z-1) + w*(5w-1), where x,y,z are nonnegative integers and w is 0 or 1.

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A306250 Number of ways to write n as x*(3x+1) + y*(3y-1) + z*(3z+2) + w*(3w-2), where x,y,z,w are nonnegative integers with x*y*z = 0.

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A069741 Let M_n be the n X n matrix M_(i,j)=1/(2^i+2^j), then a(n) is the numerator of det(M_n).

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A165410 Hankel transform of the transform of 2^n given by A165409.

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A290168 If n is even then a(n) = n^2*(n+2)/8, otherwise a(n) = (n-1)*n*(n+1)/8.

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A256560 Triangle read by rows, sums of 2 distinct nonzero squares plus sums of 2 distinct nonzero cubes: T(n,k) = n^2 + k^2 + n^3 + k^3, 1 <= k <= n-1.

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A272134 a(n) = n(15n^2 - 15*n + 4).

A306249 Number of ways to write n as x(2x-1) + y(3y-1) + z(4z-1) + w(5w-1), where x,y,z are nonnegative integers and w is 0 or 1.

A306250 Number of ways to write n as x(3x+1) + y(3y-1) + z(3z+2) + w(3w-2), where x,y,z,w are nonnegative integers with xyz = 0.

A290168 If n is even then a(n) = n^2(n+2)/8, otherwise a(n) = (n-1)n*(n+1)/8.