A129803 -id:A129803 - cp's OEIS frontend

A128862 Numbers simultaneously triangular and centered triangular.

1, 10, 136, 1891, 26335, 366796, 5108806, 71156485, 991081981, 13803991246, 192264795460, 2677903145191, 37298379237211, 519499406175760, 7235693307223426, 100780206894952201, 1403687203222107385, 19550840638214551186, 272308081731781609216

Offset: 1

Steven Schlicker, Apr 24 2007

A129803 is an essentially identical sequence. - R. J. Mathar, Jun 13 2008

			a(2)=10 because 10 is the third triangular number and the fourth centered triangular number.

Michael De Vlieger, Table of n, a(n) for n = 1..875
F. Javier de Vega, On the parabolic partitions of a number, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
J. Sadek and R. Euler, A Formula For All K-Gonal Numbers that Are Centered, 2010, arXiv:1003.2375 [math.NT], 2010.
S. C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011, pp. 339-350.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Intersection of A000217 and A005448.

Cf. A001570, A129803.

CP := n -> 1+1/2*3*(n^2-n): N:=10: u:=2: v:=1: x:=3: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+3*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp;

Rest@ CoefficientList[Series[x (1 - 5 x + x^2)/((1 - x) (1 - 14 x + x^2)), {x, 0, 19}], x] (* Michael De Vlieger, Jul 19 2023 *)

Define x(n) and y(n) by (3+sqrt(3))*(2+sqrt(3))^n = x(n) + y(n)*sqrt(3); let s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+3*(s(n)^2-s(n))).
a(n) = (3*A001570(n) + 1)/4. - Ralf Stephan, May 20 2007
From Richard Choulet, Oct 01 2007: (Start)
a(n+2) = 14*a(n+1) - a(n) - 3.
a(n+1) = 7*a(n) - 3/2 + (1/2)*sqrt(192*a(n)^2 - 96*a(n) - 15).
G.f.: x*(1-5*x+x^2)/((1-x)*(1-14*x+x^2)). (End)

Offset set to 1 by R. J. Mathar, Apr 28 2020

More terms from Michel Marcus, Jan 20 2021

A130423 Main diagonal of array A[k,n] = n-th sum of 3 consecutive k-gonal numbers, k>2.

Original entry on oeis.org

4, 14, 39, 88, 170, 294, 469, 704, 1008, 1390, 1859, 2424, 3094, 3878, 4785, 5824, 7004, 8334, 9823, 11480, 13314, 15334, 17549, 19968, 22600, 25454, 28539, 31864, 35438, 39270, 43369, 47744, 52404, 57358, 62615, 68184, 74074, 80294, 86853

Offset: 1

Jonathan Vos Post, May 26 2007

The first row of the array is the sum of 3 consecutive triangular numbers = A000217(n) + A000217(n+1) + A000217(n+2) = Centered triangular numbers: 3*n*(n-1)/2 + 1, for n>1. The second row of the array is the sum of 3 consecutive squares = Number of points on surface of square pyramid: 3*n^2 + 2 (n>1). The first column of the array is k+1 = 4, 5, 6, 7, 8, 9, ... The second column of the array is A016825 = 4*n + 2 (for n>2). The third column of the array is A017377 = 10*n + 9 (for n>0).

			The array begins:
k / A[k,n]
3.|.4.10.19.31..46..64..85.109.136.166....=A005448(n+1).
4.|.5.14.29..50..77.110.149.194.245.302...=A005918(n).
5.|.6.18.39..69.108.156.213.279.354.438...=A129863(n).
6.|.7.22.49..88.139.202.277.364.463.574...
7.|.8.26.59.107.170.248.341.449.572.710...
8.|.9.30.69.126.201.294.405.534.681.846...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Polygonal Number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A005448, A005918, A016825, A017377, A129803, A129863.

I:=[4, 14, 39, 88]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012

P := proc(k,n) n*((k-2)*n-k+4)/2 ; end: A := proc(k,n) add( P(k,i),i=n..n+2) ; end: A130423 := proc(n) A(n+3,n) ; end: seq(A130423(n),n=0..40) ; # R. J. Mathar, Jun 14 2007

CoefficientList[Series[(4-2*x+7*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)
Table[n (3n^2-3n+8)/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{4,14,39,88},40] (* Harvey P. Dale, Aug 15 2012 *)

a(n) = A[n+2,n] = P(k+2,n) + P(k+2,n+1) + P(k+2,n+2) where P(k,n) = k*((n-2)*k - (n-4))/2.
a(n) = n*(3*n^2-3*n+8)/2. G.f.: x*(4-2*x+7*x^2)/(1-x)^4. [Colin Barker, Apr 30 2012]
a(1)=4, a(2)=14, a(3)=39, a(4)=88, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Aug 15 2012

More terms from R. J. Mathar, Jun 14 2007

A133161 Indices of the triangular numbers which are also centered triangular number.

Original entry on oeis.org

1, 4, 16, 61, 229, 856, 3196, 11929, 44521, 166156, 620104, 2314261, 8636941, 32233504, 120297076, 448954801, 1675522129, 6253133716, 23337012736, 87094917229, 325042656181, 1213075707496, 4527260173804, 16895964987721

Offset: 1

Richard Choulet, Oct 09 2007

Also, indices of the triangular numbers which are sums of three consecutive triangular numbers (see A129803).

A. Kozikowska, Topological classes of statically determinate beams with arbitrary number of supports, JOURNAL OF THEORETICAL AND APPLIED MECHANICS 49, 4, pp. 1079-1100, Warsaw 2011; (see Eq. 5.18). - _N. J. A. Sloane_, Dec 17 2011.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001834, A102871, A128862, A129803, A001571.

LinearRecurrence[{5,-5,1},{1,4,16},30] (* Harvey P. Dale, Aug 29 2017 *)

a(n+2)=4*a(n+1)-a(n)+1.
a(n+1)=2*a(n)+0.5+0.5*(12*a(n)^2+12*a(n)-15)^0.5.
G.f.: x*(1-x+x^2)/(1-x)/(1-4*x+x^2). - R. J. Mathar, Oct 24 2007
a(n)-a(n-1)= A005320(n-1). - R. J. Mathar, Mar 14 2016

A129109 Sums of three consecutive hexagonal numbers.

Original entry on oeis.org

7, 22, 49, 88, 139, 202, 277, 364, 463, 574, 697, 832, 979, 1138, 1309, 1492, 1687, 1894, 2113, 2344, 2587, 2842, 3109, 3388, 3679, 3982, 4297, 4624, 4963, 5314, 5677, 6052, 6439, 6838, 7249, 7672, 8107, 8554, 9013, 9484, 9967, 10462, 10969, 11488, 12019, 12562

Offset: 0

Jonathan Vos Post, May 24 2007

Arises in hexagonal number analog to A129803 Triangular numbers which are the sum of three consecutive triangular numbers. What are the hexagonal numbers which are the sum of three consecutive hexagonal numbers? Prime for a(0) = 7, a(4) = 139, a(6) = 277, a(8) = 463, a(18) = 2113, a(22) = 3109, a(26) = 4297, a(38) = 9013, a(40) = 9967.

			a(0) = H(0) + H(1) + H(2) = 0 + 1 + 6 = 7 = 6*0^2 + 9*0 + 7.
a(1) = H(1) + H(2) + H(3) = 1 + 6 + 15 = 22 = 6*1^2 + 9*1 + 7.
a(2) = H(2) + H(3) + H(4) = 6 + 15 + 28 = 49 = 6*2^2 + 9*2 + 7.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A007667, A034961, A129803, A129863.

I:=[7,22,49]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012

LinearRecurrence[{3,-3,1},{7,22,49},50] (* Vincenzo Librandi, Feb 20 2012 *)
Total/@Partition[PolygonalNumber[6,Range[0,50]],3,1] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 14 2020 *)

a(n)=6*n^2+9*n+7 \\ Charles R Greathouse IV, Feb 20 2012

a(n) = H(n) + H(n+1) + H(n+2) where H(n) = A000384(n) = n*(2*n-1).
a(n) = 6*n^2 + 9*n + 7.
From Colin Barker, Feb 20 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (7 + x + 4*x^2)/(1-x)^3. (End)
E.g.f.: (7 + 15*x + 6*x^2)*exp(x). - Elmo R. Oliveira, Nov 16 2024

A129111 Sums of three consecutive heptagonal numbers.

Original entry on oeis.org

8, 26, 59, 107, 170, 248, 341, 449, 572, 710, 863, 1031, 1214, 1412, 1625, 1853, 2096, 2354, 2627, 2915, 3218, 3536, 3869, 4217, 4580, 4958, 5351, 5759, 6182, 6620, 7073, 7541, 8024, 8522, 9035, 9563, 10106, 10664, 11237, 11825, 12428, 13046, 13679, 14327, 14990

Offset: 0

Jonathan Vos Post, May 24 2007

Arises in heptagonal number analog to A129803 (Triangular numbers which are the sum of three consecutive triangular numbers).
What are the heptagonal numbers which are the sum of three consecutive heptagonal numbers?
Prime for a(2) = 59, a(3) = 107, a(7) = 449, a(10) = 863, a(11) = 1031, a(23) = 4217, a(26) = 5351, a(31) = 7541, a(42) = 13679, a(43) = 14327, a(46) = 16361, a(51) = 20051.

			a(0) = Hep(0) + Hep(1) + Hep(2) = 0 + 1 + 7 = 8 = (15/2)*0^2 + (21/2)*0 + 8.
a(1) = Hep(1) + Hep(2) + Hep(3) = 1 + 7 + 18 = 26 = (15/2)*1^2 + (21/2)*1 + 8.
a(2) = Hep(2) + Hep(3) + Hep(4) = 7 + 18 + 34 = 59 = (15/2)*2^2 + (21/2)*2 + 8.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000566, A007667, A034961, A129803, A129863.

I:=[8,26,59]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012

LinearRecurrence[{3,-3,1},{8,26,59},50] (* Vincenzo Librandi, Feb 12 2012 *)

a(n)=3*n*(5*n+7)/2+8 \\ Charles R Greathouse IV, Jun 17 2017

def a(n): return 3*n*(5*n+7)//2 + 8
print([a(n) for n in range(44)]) # Michael S. Branicky, Aug 26 2021

a(n) = Hep(n) + Hep(n+1) + Hep(n+2) where Hep(n) = A000566(n) = n*(5*n-3)/2.
a(n) = (15/2)*n^2 + (21/2)*n + 8.
From Colin Barker, Feb 20 2012: (Start)
G.f.: (8 + 2*x + 5*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
E.g.f.: exp(x)*(16 + 36*x + 15*x^2)/2. - Elmo R. Oliveira, Nov 16 2024

A130424 Main diagonal of array A[k,n] = n-th sum of k consecutive k-gonal numbers, k>2.

Original entry on oeis.org

4, 30, 125, 365, 854, 1724, 3135, 5275, 8360, 12634, 18369, 25865, 35450, 47480, 62339, 80439, 102220, 128150, 158725, 194469, 235934, 283700, 338375, 400595, 471024, 550354, 639305, 738625, 849090, 971504, 1106699, 1255535, 1418900

Offset: 1

Jonathan Vos Post, May 26 2007

The first row of the array is the sum of 3 consecutive triangular numbers = A000217(n) + A000217(n+1) + A000217(n+2) = Centered triangular numbers: 3*n*(n-1)/2 + 1, for n>1. The second row of the array is the sum of 4 consecutive squares = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = A027575(n). The third row of the array is the sum of 5 consecutive pentagonal numbers.

			The array begins:
k / A[k,n]
3.|...4..10..19...31...46...64...85..109.136.166...=A005448(n+1).
4.|..14..30..54...86..126..174..230..294.366.446...=A027575(n).
5.|..40..75.125..190..270..365..475..600.740...
6.|..95.161.251..365..503..665..851.1061.1295...
7.|.196.308.455..637..854.1106.1393.1715.2072...
8.|.364.540.764.1036.1356.1724.2140.2604.3116...

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Polygonal Number.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A005448, A005918, A016825, A017377, A129803, A129863, A130423.

P := proc(k,n) n*((k-2)*n-k+4)/2 ; end: A := proc(k,n) add( P(k,i),i=n..n+k-1) ; end: A130424 := proc(n) A(n+3,n) ; end: seq(A130424(n),n=0..40) ; # R. J. Mathar, Oct 28 2007

LinearRecurrence[{5,-10,10,-5,1},{4,30,125,365,854},50] (* Harvey P. Dale, Jun 23 2020 *)

a(n) = A[n+2,n] = P(k+2,n) + P(k+2,n+1) + P(k+2,n+2) + ... P(k+2,n+k-1) where P(k,n) = k*((n-2)*k - (n-4))/2.
a(n) = (n+2)*(7*n^3-8*n^2+12*n-3)/6. [R. J. Mathar, Oct 30 2008]
G.f.: x*(4+10*x+15*x^2-x^4)/(1-x)^5. [Colin Barker, Sep 08 2012]

More terms from R. J. Mathar, Oct 28 2007

A238017 Least triangular number representable as a sum of n consecutive triangular numbers, or -1 if no such triangular number exists.

Original entry on oeis.org

0, 1, 10, 10, 55, -1, 210, 120, 120, 1485, 2145, -1, 2080, -1, -1, 56616, 1326, 12561, -1, 1540, 1540, 21736, -1, -1, 52650, 16653, 4950, 26796, 10440, 12880, 7750, -1, -1, 7140, 7140, 154290, -1, 11476, -1, 214840, -1, -1, 207690, 23252790, -1, -1, 6895041, -1, 750925

Offset: 1

Alex Ratushnyak, Feb 17 2014

sign

			a(5) = 55 because 55 is the least triangular number representable as a sum of five consecutive triangular numbers: 55 = 3 + 6 + 10 + 15 + 21.
a(7) = 210 because 210 is the least triangular number representable as a sum of seven consecutive triangular numbers: 210 = 10 + 15 + 21 + 28 + 36 + 45 + 55.
10 appears twice because 10 = 1 + 3 + 6 and 10 = 0 + 1 + 3 + 6.

Cf. A000217, A129803, A131557, A238018.

a[1] = 0; a[n_] := Block[{t, x, y, s = Reduce[n*(-1+3*t^2+3*t*n+n^2)/6 == x*(x+1)/2 && x>0 && t >= 0, {t, x}, Integers]}, If[s === False, -1, y = Min[x /. List @ ToRules @ Expand[s /. C[1] -> 1]]; y*(y+1)/2]]; Array[a, 49] (* Giovanni Resta, Mar 02 2014 *)

a(6) and a(12)-a(49) from Jon E. Schoenfield and Giovanni Resta, Mar 04 2014

A238018 Least k such that the sum triangular(k) + triangular(k+1) +...+ triangular(k+n-1) is a triangular number, or -1 if no such k exists.

Original entry on oeis.org

0, 0, 1, 0, 2, -1, 4, 1, 0, 12, 14, -1, 11, -1, -1, 76, 3, 28, -1, 1, 0, 33, -1, -1, 52, 22, 4, 29, 11, 13, 5, -1, -1, 1, 0, 74, -1, 3, -1, 83, -1, -1, 76, 1006, -1, -1, 518, -1, 150, -1, -1, 103, 133, 51, 14, 45, 19, -1, 5, -1

Offset: 1

Alex Ratushnyak, Feb 17 2014

sign

			a(5) = 2 because 2 is the least integer such that the sum of 5 consecutive triangular numbers starting with triangular(2) is a triangular number: 55 = 3+6+10+15+21.
a(7) = 4 because 4 is the least integer such that the sum of 7 consecutive triangular numbers starting with triangular(4) is a triangular number: 210 = 10+15+21+28+36+45+55.

Cf. A000217, A129803, A238017.

s[n_, k_] := n*(3*k^2 + 3*k*n + n^2 - 1)/6; a[1] = 0; a[n_] := Module[{r}, r = Reduce[ k >= 0 && m >= 0 && 8*s[n, k] + 1 == m^2 , {k, m}, Integers] /. C[1] -> 1 // FullSimplify; If[r === False, -1, k /. {ToRules[r]} // Min]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Feb 26 2014 *)

More terms from Jean-François Alcover, Feb 26 2014

cp's OEIS Frontend

A128862 Numbers simultaneously triangular and centered triangular.

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A130423 Main diagonal of array A[k,n] = n-th sum of 3 consecutive k-gonal numbers, k>2.

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A133161 Indices of the triangular numbers which are also centered triangular number.

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A129109 Sums of three consecutive hexagonal numbers.

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A129111 Sums of three consecutive heptagonal numbers.

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A130424 Main diagonal of array A[k,n] = n-th sum of k consecutive k-gonal numbers, k>2.

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A238017 Least triangular number representable as a sum of n consecutive triangular numbers, or -1 if no such triangular number exists.

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