A154293 -id:A154293 - cp's OEIS frontend

A154297 Primes of the form (1+2+3+...+m)/21 = A000217(m)/21, for some m.

5, 11, 41, 43

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

This asks for primes p which are a triangular number divided by 21, or, 2*3*7*p=k*(k+1) for some k. Matching factors shows that the sequence is complete [R. J. Mathar, Aug 15 2010]
Original definition: Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=21.
The corresponding m-values are m=14, 21, 41, 42 (cf. A154296).  It is clear that for m>42, A000217(m)/21 = m(m+1)/42 cannot be a prime. - M. F. Hasler, Dec 31 2012

Cf. A057570, A154293, A154296, A154298, A154299, A154300, A154301, A154302, A154303, A154304.

lst={};s=0;Do[s+=n/21;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,9!}];lst
#/21&/@Select[Accumulate[Range[100]],PrimeQ[#/21]&] (* Harvey P. Dale, Dec 17 2012 *)

select(x->denominator(x)==1 & isprime(x), vector(42, m, m^2+m)/42)  \\ - M. F. Hasler, Dec 31 2012

Added keywords fini,full - R. J. Mathar, Aug 15 2010

Edited by M. F. Hasler, Dec 31 2012

A154300 Primes of the form (1+2+...+m)/57 = A000217(m)/57.

Original entry on oeis.org

3, 13, 29, 113

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Original definition: "Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=57."
Primes which are some triangular number divided by 57. Finiteness of the sequence follows along the reasoning in A154297.
The corresponding m-values are m=18,38,57,113. It is clear that for m>2*57, T(m)/57 = m(m+1)/114 cannot be a prime, since then each factor in the numerator is larger than the denominator. See A154304 for further comments and PARI code. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/57;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,2*9!}];lst
Select[Accumulate[Range[1000]]/57,PrimeQ] (* Harvey P. Dale, Jun 24 2015 *)

d=57*2;for(m=1,999,(m^2+m)%d==0&isprime((m^2+m)/d)&print1(m",")) \\ print the m-values(!) - use A154304(57) to get A154300 as a vector. \\ - M. F. Hasler, Jan 06 2013

Keywords fini,full added by R. J. Mathar, Aug 15 2010

Edited by M. F. Hasler, Jan 06 2013

A154301 Primes of the form (1+2+...+m)/75 = A000217(m)/75.

Original entry on oeis.org

17, 37, 149, 151

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Original definition: "Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=75."

The corresponding m-values are m=50,74,149,150. It is clear that for m>2*75, T(m)/75 = m(m+1)/150 cannot be a prime, since then each factor in the numerator is larger than the denominator. See A154304 for further comments and PARI code. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/75;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,3*9!}];lst

d=75*2;for(m=1,999,(m^2+m)%d==0&isprime((m^2+m)/d)&print1(m",")) \\ print the m-values(!) - use A154304(75) to get A154301 as a vector. \\ - M. F. Hasler, Jan 06 2013

Keywords fini,full added by R. J. Mathar, Aug 15 2010

Edited by M. F. Hasler, Jan 06 2013

A154302 Primes of the form (1+2+...+m)/87 = A000217(m)/87.

Original entry on oeis.org

5, 19, 43, 173

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Original definition: "Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=87."

The corresponding m-values are m=29,57,86,173. It is clear that for m>2*87, T(m)/87 = m(m+1)/174 cannot be a prime, since then each factor in the numerator is larger than the denominator. See A154304 for further comments and PARI code. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/87;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,4*9!}];lst

d=87*2;for(m=1,999,(m^2+m)%d==0&isprime((m^2+m)/d)&print1(m",")) \\ print the m-values(!) - use A154304(87) to get A154302 as a vector. \\ - M. F. Hasler, Jan 06 2013

Edited by M. F. Hasler, Jan 06 2013

A164578 Integers of the form (k+1)*(2k+1)/12.

Original entry on oeis.org

10, 23, 65, 94, 168, 213, 319, 380, 518, 595, 765, 858, 1060, 1169, 1403, 1528, 1794, 1935, 2233, 2390, 2720, 2893, 3255, 3444, 3838, 4043, 4469, 4690, 5148, 5385, 5875, 6128, 6650, 6919, 7473, 7758, 8344, 8645, 9263, 9580, 10230, 10563, 11245, 11594

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

This can also be defined as integer averages of the first k halved squares, 1^2/2, 2^2/2, 3^2/2,... , 3^k/2, because sum_{j=1..k} j^2/2 = k*(k+1)*(2k+1)/12. The generating k are in A168489.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A078617, A078618, A154293, A164576, A164577.

s=0;lst={};Do[a=(s+=(n^2)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,2*6!}];lst
Select[Table[((n+1)(2n+1))/12,{n,300}],IntegerQ] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{10,23,65,94,168},60] (* Harvey P. Dale, Jun 14 2017 *)

Vec(x*(10+13*x+22*x^2+3*x^3)/((1-x)^3*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 26 2016

a(n) = +a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5). G.f. x*(-10-13*x-22*x^2-3*x^3) / ((1+x)^2*(x-1)^3). - R. J. Mathar, Jan 25 2011
From Colin Barker, Jan 26 2016: (Start)
a(n) = (24*n^2+6*n-(-1)^n*(8*n+1)+1)/4.
a(n) = (12*n^2-n)/2 for n even.
a(n) = (12*n^2+7*n+1)/2 for n odd.
(End)

A193828 Even generalized pentagonal numbers.

Original entry on oeis.org

0, 2, 12, 22, 26, 40, 70, 92, 100, 126, 176, 210, 222, 260, 330, 376, 392, 442, 532, 590, 610, 672, 782, 852, 876, 950, 1080, 1162, 1190, 1276, 1426, 1520, 1552, 1650, 1820, 1926, 1962, 2072, 2262, 2380, 2420, 2542, 2752, 2882, 2926, 3060, 3290, 3432, 3480

Offset: 0

Omar E. Pol, Aug 19 2011

Even numbers in A001318.

Exponents in the expansion of Sum_{n >= 0} q^(2*n)/(Product_{k = 1..2*n} 1 + q^(2*k)) = 1 + q^2 - q^12 - q^22 + q^26 + q^40 - - + + ... (follows from Berndt et al., Theorem 3.3). Cf. A067589. - Peter Bala, Jan 21 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
B. C. Berndt, B. Kim, and A. J. Yee, Ramanujan's lost notebook: Combinatorial proofs of identities associated with Heine's transformation or partial theta functions, J. Comb. Thy. Ser. A, 117 (2010), 957-973.
Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157 (December 2015), Pages 223-232.
Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Cf. A001318, A067589, A154293.

CoefficientList[Series[-2*x*(x^2 - x + 1)*(x^2 + 4*x + 1)/((x - 1)^3*(x^2 + 1)^2), {x, 0, 50}], x] (* G. C. Greubel, Jun 06 2017 *)
LinearRecurrence[{3,-5,7,-7,5,-3,1},{0,2,12,22,26,40,70},50] (* Harvey P. Dale, Apr 09 2019 *)

my(x='x+O('x^50)); concat([0], Vec(-2*x*(x^2-x+1)*(x^2+4*x+1)/((x-1)^3*(x^2+1)^2))) \\ G. C. Greubel, Jun 06 2017

a(n) = A000217(A108752(n+1))/3 = 2*A154293(n+1).
G.f.: -2*x*(x^2-x+1)*(x^2+4*x+1)/((x-1)^3*(x^2+1)^2). - Colin Barker, Sep 12 2012
Sum_{n>=1} 1/a(n) = 6 - (1+4/sqrt(3))*Pi/2. - Amiram Eldar, Mar 18 2022

A154298 Primes of the form (1+...+m)/33 = A000217(m)/33, for some m.

Original entry on oeis.org

2, 7, 17, 67

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Primes which are some triangular number A000217 divided by 33. Finiteness is shown with the same strategy as in A154297.
Original definition: Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=33.
The corresponding m-values are m=11, 21, 33, 66 (cf. A154296).  It is clear that for m>66, A000217(m)/33 = m(m+1)/66 cannot be a prime. - M. F. Hasler, Dec 31 2012

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/33;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,9!}];lst
Select[Accumulate[Range[200]]/33,PrimeQ] (* Harvey P. Dale, Aug 11 2025 *)

select(x->denominator(x)==1 & isprime(x), vector(66, m, m^2+m)/66)  \\ - M. F. Hasler, Dec 31 2012

Edited by M. F. Hasler, Dec 31 2012

A154299 Primes of the form (1+...+m)/51 = A000217(m)/51.

Original entry on oeis.org

3, 11, 101, 103

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Primes p of the form k*(k+1)/(2*51): There is a finite set of solutions to 2*3*17*p=k*(k+1). - R. J. Mathar, Aug 15 2010
Original definition: "Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=51."
The corresponding m-values are m=17, 33, 101, 102 (cf. A154296-A154304, see the latter for more comments and PARI code).  It is clear that for m>102, A000217(m)/51 = m(m+1)/102 has at least 2 factors and hence cannot be prime. - M. F. Hasler, Dec 31 2012

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/51;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,9!}];lst
Select[Accumulate[Range[1000]/51],PrimeQ] (* Harvey P. Dale, Jun 21 2012 *)

M=51*2;select(x->denominator(x)==1 & isprime(x), vector(M, m, m^2+m)/M)  \\ M. F. Hasler, Dec 31 2012

Keywords fini,full added by R. J. Mathar, Aug 15 2010

A164579 Integer averages of halves of first cubes of natural numbers (n^3)/2 for some n.

Original entry on oeis.org

6, 56, 81, 198, 480, 578, 950, 1656, 1875, 2646, 3968, 4356, 5670, 7800, 8405, 10406, 13536, 14406, 17238, 21560, 22743, 26550, 32256, 33800, 38726, 46008, 47961, 54150, 63200, 65610, 73206, 84216, 87131, 96278, 109440, 112908, 123750, 139256

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Also, integers of the form (1/8)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009

			1/2, 9/4, 6, 25/2, 45/2, 147/4, 56, 81, ...

Index entries for linear recurrences with constant coefficients, signature (1,0,3,-3,0,-3,3,0,1,-1).

Cf. A050248, A051456, A078617, A078618, A136116, A154293, A164576, A164577, A164578

s=0;lst={};Do[a=(s+=(n^3)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,3*5!}];lst
LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{6,56,81,198,480,578,950,1656,1875,2646},40] (* Harvey P. Dale, Jul 26 2017 *)
Module[{nn=200,ac},ac=Accumulate[Range[nn]^3/2];Select[#[[1]]/#[[2]]&/@ Thread[{ac,Range[nn]}],IntegerQ]] (* Harvey P. Dale, Jan 28 2020 *)

forstep(n=3, 150, [4,1,3], print1(n*(n+1)^2>>3, ", ")); \\ Charles R Greathouse IV, Nov 02 2009

G.f.: ( x*(6+50*x+25*x^2+99*x^3+132*x^4+23*x^5+39*x^6+10*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011

A180926 Numbers k such that 6k and 10k are triangular numbers.

Original entry on oeis.org

0, 1, 63, 3906, 242110, 15006915, 930186621, 57656563588, 3573776755836, 221516502298245, 13730449365735355, 851066344173293766, 52752382889378478138, 3269796672797292350791, 202674641330542747270905

Offset: 1

Vladimir Pletser, Sep 25 2010

From Klaus Purath, Jul 25 2024: (Start)
Numbers k such that 48k + 1 is a square as well as the sum of two consecutive terms.
a(n) = t(n-1)*t(n)/(8*t(1)^2) where (t) is any recurrence t(k) = 8*t(k-1) - t(k-2) with t(0) = 0 and arbitrary t(1) != 0. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Subsequence of A154293.

Cf. A069498, A154293.

a[1] = 0; a[n_] := a[n] = (62 a[n - 1] + 1 + Sqrt[(48 a[n - 1] + 1)*(80 a[n - 1] + 1)])/2; Array[a, 14] (* Robert G. Wilson v, Sep 27 2010 *)
Rest[CoefficientList[Series[-x^2/((x - 1) (x^2 - 62 x + 1)), {x, 0, 30}], x]] (* Vincenzo Librandi, Jun 26 2014 *)
LinearRecurrence[{63,-63,1},{0,1,63},20] (* Harvey P. Dale, Dec 25 2019 *)

isok(n) = ispolygonal(6*n, 3) && ispolygonal(10*n, 3); \\ Michel Marcus, Jun 25 2014

a(n) = (62*a(n-1) + 1 + ((48*a(n-1) + 1)*(80*a(n-1) + 1))^(1/2))/2 with a(1)=0.
G.f.: -x^2 / ((x-1)*(x^2-62*x+1)). - Colin Barker, Jun 25 2014
a(n) = (-8+(4+sqrt(15))*(31+8*sqrt(15))^(-n) - (-4+sqrt(15))*(31+8*sqrt(15))^n)/480. - Colin Barker, Mar 03 2016

a(8) onwards from Robert G. Wilson v, Sep 27 2010

cp's OEIS Frontend

A154297 Primes of the form (1+2+3+...+m)/21 = A000217(m)/21, for some m.

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A154300 Primes of the form (1+2+...+m)/57 = A000217(m)/57.

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A154301 Primes of the form (1+2+...+m)/75 = A000217(m)/75.

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A154302 Primes of the form (1+2+...+m)/87 = A000217(m)/87.

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A164578 Integers of the form (k+1)*(2k+1)/12.

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A193828 Even generalized pentagonal numbers.

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A154298 Primes of the form (1+...+m)/33 = A000217(m)/33, for some m.

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A154299 Primes of the form (1+...+m)/51 = A000217(m)/51.

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A180926 Numbers k such that 6k and 10k are triangular numbers.