A057570 -id:A057570 - 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

A346798 Number of partitions of n into parts congruent to 0, 3 or 4 (mod 7).

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 1, 2, 1, 1, 3, 3, 2, 3, 6, 4, 4, 8, 9, 6, 10, 15, 12, 12, 21, 22, 18, 25, 36, 30, 32, 48, 52, 45, 60, 78, 72, 75, 105, 113, 105, 130, 166, 156, 166, 218, 236, 224, 274, 332, 325, 345, 436, 469, 462, 544, 649, 644, 688, 839, 907, 903, 1051

Offset: 0

Ludovic Schwob, Aug 04 2021

nonn

			For n=19 the a(19)=6 solutions are 3+3+3+3+3+4, 3+3+3+3+7, 3+3+3+10, 3+4+4+4+4, 4+4+4+7, and 4+4+11.

Ludovic Schwob, Table of n, a(n) for n = 0..10000

Cf. A000041, A047342, A057570, A006950, A036820, A036822, A195848, A035363, A195849, A346797.

CoefficientList[Series[Product[1/((1 - x^(7*k))(1 - x^(7*k-3))(1 - x^(7*k-4))),{k,55}],{x,0,55}],x] (* Stefano Spezia, Aug 04 2021 *)

G.f.: Product_{k>=1} 1/((1 - x^(7*k))*(1 - x^(7*k-3))*(1 - x^(7*k-4))).
a(n) = a(n-3) + a(n-4) - a(n-13) - a(n-15) + + - - (with a(0)=1 and a(n) = 0 for negative n), where 3, 4, 13, 15, ... is the sequence A057570.
a(n) ~ exp(Pi*sqrt(2*n/7)) / (8*cos(Pi/14)*n). - Vaclav Kotesovec, Aug 05 2021

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

A193470 Square array A(n,k) (n>=1, k>=0) read by antidiagonals: A(n,0) = 0 and A(n,k) is the least integer > A(n,k-1) that can be expressed as a triangular number divided by n.

Original entry on oeis.org

0, 0, 1, 0, 3, 3, 0, 1, 5, 6, 0, 7, 2, 14, 10, 0, 2, 9, 5, 18, 15, 0, 1, 3, 30, 7, 33, 21, 0, 3, 6, 9, 34, 12, 39, 28, 0, 15, 4, 11, 11, 69, 15, 60, 36, 0, 4, 17, 13, 13, 21, 75, 22, 68, 45, 0, 1, 5, 62, 15, 20, 24, 124, 26, 95, 55, 0, 5, 12, 17, 66, 30, 35, 38, 132, 35, 105, 66

Offset: 1

Peter Luschny, Jul 27 2011

			n\k  0   1   2    3    4     5     6     7
------------------------------------------
1 |  0   1   3    6   10    15    21    28    A000217
2 |  0   3   5   14   18    33    39    60    A074378
3 |  0   1   2    5    7    12    15    22    A001318
4 |  0   7   9   30   34    69    75   124    A154260
5 |  0   2   3    9   11    21    24    38    A057569
6 |  0   1   6   11   13    20    35    46    A154293
7 |  0   3   4   13   15    30    33    54    A057570
8 |  0  15  17   62   66   141   147   252    A157716

Cf. A061782, A011772.

A193470_rect := proc(n,k) local j,i,L; L := NULL; j := 0; while nops([L]) < k do add(i/n, i=1..j); if type(%,integer) then L := L,% fi; j := j+1 od; L end:
seq(print(A193470_rect(n, 12)),n = 1..8);

a[, 0] = 0; a[n, k_] := a[n, k] = For[j = a[n, k-1]+1, True, j++, If[Reduce[m > 0 && j == m(m+1)/(2n), m, Integers] =!= False, Return[j]]]; Table[a[n-k, k], {n, 1, 12}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 07 2016 *)

A364554 a(n) = number of primes of the form T(k)/n, for some k, where T(k)=A000217(k) is a triangular number.

Original entry on oeis.org

1, 2, 3, 1, 3, 2, 2, 1, 3, 1, 2, 2, 1, 2, 4, 1, 0, 2, 1, 1, 4, 2, 2, 2, 1, 2, 2, 0, 1, 3, 1, 0, 4, 1, 3, 2, 2, 1, 3, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 1, 4, 1, 1, 3, 1, 1, 4, 1, 1, 3, 1, 0, 2, 1, 2, 1, 0, 2, 2, 2, 0, 0, 1, 1, 4, 1, 1, 2, 1, 0, 2, 1, 2, 2, 2, 2, 4, 0, 1, 4

Offset: 1

Lamine Ngom, Jul 28 2023

nonn

Implementing a suggestion in the comment section of sequences A154296, ..., A154304, this sequence computes the number of primes of the form T(k)/n.
Equivalently, number of primes p such that 8*n*p+1 is a perfect square.
Let's consider, for all primes p, the set of linear recurrences {b(m)} defined as follows:
If p = 2, then {b(m)} = A074378 (numbers of the form x*(4*x -+ 1)); otherwise, b(m) = b(m-1) + 2*b(m-2) - 2*b(m-3) - b(m-4) + b(m-5) with initial terms b(0) = 0, b(1) = (p-1)/2, b(2) = (p+1)/2, b(3) = 2*p-1 and b(4) = 2*p+1. Numbers of the form x*(p*x -+ 1)/2.
Then a(n) = number of sequences {b(m)} in which n is a term.
This implies that:
i) for any n, the largest prime of the form T(k)/n is at most 2*n+1;
ii) if n is prime, then a(n) < 4. (3 and 5 are the only primes p such that a(p) = 3; primes p such that a(p) = 0 are A109998.)
Deeper in the examination of these results, we notice that the set of primes p of the form T(k)/n arises from the factorization of n. This set is exactly all primes p of the form (2*r -+ 1)/d or (r -+ 1)/(2*d), where d is some divisor of n and r is the ratio n/d. (Proof is welcome.)
Indices k of corresponding triangular numbers T(k) such that T(k) = n*p are then:
  2*r     if p = (2*r + 1)/d,
  2*r - 1 if p = (2*r - 1)/d,
    r     if p = (r + 1)/(2*d),
    r - 1 if p = (r - 1)/(2*d).
And pluging the value of p in the equivalent definition, the expression 8*n*p+1 yields respectively to following perfect squares: (4*r+1)^2, (4*r-1)^2, (2*r+1)^2 and (2*r-1)^2.

			a(15) = 4 since there are exactly 4 triangular numbers T(k) such that T(k) = 15*p, with p prime.
T(9)/15 = 45/15 = 3, T(14)/15 = 105/15 = 7, T(29)/15 = 435/15 = 29 and T(30)/15 = 465/15 = 31.
a(17) = 0 since there is no triangular number T(k) such that T(k) = 17*p, with p prime.

Cf. A000217, A154296, ..., A154304, A074378, A001318, A057569, A057570, A109998.

Cf. A364555 (indices of 0's).

Conjecture: a(n) = number of primes in the union of sets {(2*r -+ 1)/d; (r -+ 1)/(2*d)}, with d divisor of n and r = n/d.

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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A346798 Number of partitions of n into parts congruent to 0, 3 or 4 (mod 7).

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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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A193470 Square array A(n,k) (n>=1, k>=0) read by antidiagonals: A(n,0) = 0 and A(n,k) is the least integer > A(n,k-1) that can be expressed as a triangular number divided by n.

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