A154293 -id:A154293 - cp's OEIS frontend

A287746 First differences of A154293.

1, 5, 5, 2, 7, 15, 11, 4, 13, 25, 17, 6, 19, 35, 23, 8, 25, 45, 29, 10, 31, 55, 35, 12, 37, 65, 41, 14, 43, 75, 47, 16, 49, 85, 53, 18, 55, 95, 59, 20, 61, 105, 65, 22, 67, 115, 71, 24, 73, 125, 77, 26, 79, 135, 83, 28, 85, 145, 89, 30, 91, 155, 95, 32, 97, 165, 101, 34, 103, 175

Offset: 1

Greg Huber, May 31 2017

nonn

Cf. A069497, A108782, A154293, A287733.

Select[Accumulate[Range[0,300]]/6,IntegerQ]//Differences (* Harvey P. Dale, Feb 20 2020 *)

a(n) = A287733(n)/6.

A057570 Numbers of the form n*(7n+-1)/2.

Original entry on oeis.org

0, 3, 4, 13, 15, 30, 33, 54, 58, 85, 90, 123, 129, 168, 175, 220, 228, 279, 288, 345, 355, 418, 429, 498, 510, 585, 598, 679, 693, 780, 795, 888, 904, 1003, 1020, 1125, 1143, 1254, 1273, 1390, 1410, 1533, 1554, 1683, 1705, 1840, 1863, 2004

Offset: 1

N. J. A. Sloane, Oct 04 2000

Also integers of the form Sum_{k = 1..n} k/7. - Alonso del Arte, Jan 20 2012
Sequence provides all integers m such that 56*m + 1 is a square. [Bruno Berselli, Oct 07 2015]
The sequence terms occur as the exponents in the expansion of Product_{n >= 1} (1 - x^(7*n)) * (1 + x^(7*n-3)) * (1 + x^(7*n-4)) = 1 + x^3 + x^4 + x^13 + x^15 + x^30 + x^33 + .... Cf. A363801. - Peter Bala, Nov 21 2024

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

Cf. A074378, A001318, A057569, A154260, A154292, A154293, A047341, A274830, A363801.

Select[Table[Plus@@Range[n]/7, {n, 0, 199}], IntegerQ] (* Alonso del Arte, Jan 20 2012 *)
CoefficientList[Series[-x (3 + x + 3 x^2) / ((1 + x)^2 (x - 1)^3), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 19 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,3,4,13,15},50] (* Harvey P. Dale, Sep 17 2023 *)

a(n)=(14*n*(n-1)+5*(2*n-1)*(-1)^n+5)/16 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: -x^2*(3+x+3*x^2) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Jan 25 2011
a(n) = +1*a(n-1)+2*a(n-2)-2*a(n-3)-1*a(n-4)+1*a(n-5). - Joerg Arndt, Jan 25 2011
a(n) = (14*n*(n-1)+5*(2*n-1)*(-1)^n+5)/16. - Bruno Berselli, Jan 25 2011
a(n)-a(n-2) = A047341(n-1) for n>2. - Bruno Berselli, Jan 25 2011
Sum_{n>=2} 1/a(n) = 14 - 2*cot(Pi/7)*Pi. - Amiram Eldar, Mar 17 2022

A154296 Primes of the form (1+2+3+...+m)/15 = A000217(m)/15, for some m.

Original entry on oeis.org

3, 7, 29, 31

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=15.
The corresponding m-values are m=9, 14, 29, 30. It is clear that for m > 30, T(m)/15 = m*(m+1)/30 cannot be a prime. - M. F. Hasler, Dec 31 2012
All of the sequences A154296, ..., A154304 could or should be grouped together in a single ("fuzzy"?) table. It would be more interesting to have the function f(n) which gives the *number* of primes of the form T(k)/n. - M. F. Hasler, Jan 06 2013
Also primes p such that 120*p+1 is a perfect square. - Lamine Ngom, Jul 22 2023

Cf. A057570, A154293, A154297, A154298, A154299, A154300, A154301, A154302, A154303, A154304.

lst={};s=0;Do[s+=n/15;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,9!}];lst
Select[(Accumulate[Range[200]])/15,PrimeQ] (* Harvey P. Dale, Oct 30 2011 *)

select(x->denominator(x)==1 & isprime(x), vector(30,m,m^2+m)/30)  \\ M. F. Hasler, Dec 31 2012

Edited by M. F. Hasler, Dec 31 2012

A154304 Primes of the form (1+2+...+m)/210 = A000217(m)/210.

Original entry on oeis.org

3, 17, 47, 419, 421

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=210.

The corresponding m-values are m=35,84,140,419,420. It is clear that for m>420, T(m)/210 = m(m+1)/420 cannot be a prime, since then each factor in the numerator is larger than the denominator. All of the sequences A154296, ..., A154304 could or should be grouped together in a single ("fuzzy"?) table. It would be more interesting to have the function f(n) which gives the *number* of primes of the form T(k)/n. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154303.

lst={};s=0;Do[s+=n/210;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,6*9!}];lst

A154304(d=210)={select(x->denominator(x)==1 && isprime(x), vector(d*=2, m, m^2+m)/d)}  \\ - M. F. Hasler, Jan 06 2013

Edited by M. F. Hasler, Jan 06 2013

A164576 Integer averages of the set of the first positive squares up to some n^2.

Original entry on oeis.org

1, 11, 20, 46, 63, 105, 130, 188, 221, 295, 336, 426, 475, 581, 638, 760, 825, 963, 1036, 1190, 1271, 1441, 1530, 1716, 1813, 2015, 2120, 2338, 2451, 2685, 2806, 3056, 3185, 3451, 3588, 3870, 4015, 4313, 4466, 4780, 4941, 5271, 5440, 5786, 5963, 6325, 6510

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Integers of the form A000330(k)/k, k listed in A007310. - R. J. Mathar, Aug 20 2009

			a(1) = 1^2/1 is an integer. The average of the first two squares is (1^2+2^2)/2=5/2, not integer.
The average of the first three squares is (1^2+2^2+3^2)/3=14/3, not integer.
The average of the first five squares is (1^2+2^2+3^2+4^2+5^2)/ 5=11, integer, and constitutes a(2).

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

Cf. A050248, A078617, A154293.

s=0;lst={};Do[a=(s+=n^2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,6!}];lst
Flatten[Table[{(1 + 3 k) (1 + 4 k), (1 + k) (11 + 12 k)}, {k, 0, 499}]] (* Zak Seidov, Aug 15 2012 *)
Module[{nn=150,sq},sq=Range[nn]^2;Select[Table[Mean[Take[sq,n]],{n,nn}],IntegerQ]] (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,11,20,46,63},50] (* Harvey P. Dale, Oct 31 2013 *)

a(n) = 1/4*(12*n^2 - 6*n + (-1)^n*(4*n-1) + 1) \\ Colin Barker, Dec 26 2015

a(n) = A000330(A007310(n)) / A007310(n) = A175485(A007310(n)). - Jaroslav Krizek, May 28 2010
G.f. ( -x*(1+10*x+7*x^2+6*x^3) ) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Jan 25 2011
a(n) = 1/4*(12*n^2 - 6*n + (-1)^n*(4*n-1) + 1). - Colin Barker, Dec 26 2015

Edited by R. J. Mathar, Aug 20 2009

A067589 Numbers k such that A067588(k) is an odd number.

Original entry on oeis.org

1, 5, 7, 15, 35, 51, 57, 77, 117, 145, 155, 187, 247, 287, 301, 345, 425, 477, 495, 551, 651, 715, 737, 805, 925, 1001, 1027, 1107, 1247, 1335, 1365, 1457, 1617, 1717, 1751, 1855, 2035, 2147, 2185, 2301, 2501, 2625, 2667, 2795, 3015, 3151, 3197, 3337

Offset: 1

Naohiro Nomoto, Jan 31 2002

The terms are exactly the odd pentagonal numbers; that is, they are all the odd numbers of the form k*(3*k-1)/2 where k is an integer. - James Sellers, Jun 09 2007
Apparently groups of two odd pentagonal numbers (A000326, A014632) followed by two odd 2nd pentagonal numbers (A005449), which leads to the conjectured generating function x*(x^2+4*x+1)*(x^4-2*x^3+4*x^2-2*x+1)/((x^2+1)^2*(1-x)^3). - R. J. Mathar, Jul 26 2009
Odd generalized pentagonal numbers. - Omar E. Pol, Aug 19 2011
From Peter Bala, Jan 10 2025: (Start)
The sequence terms are the exponents in the expansion of Sum_{n >= 0} x^(2*n+1)/(Product_{k = 1..2*n+1} 1 + x^(2*k+1)) = x + x^5 - x^7 - x^15 + x^35 + x^51 - x^57 - x^77 + + - - ... (follows from Berndt et al., Theorem 3.3). Cf. A193828.
For positive integer m, define b_m(n) = Sum_{k = 1..n} k^(2*m+1)*A000009(k)*A000009(n-k). We conjecture that
i) for odd n, b(n)/ n is an integer
ii) b(2*n)/n is an integer, which is odd iff n is a member of this sequence.
Cf. A067567. (End)

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

Cf. A067567, A067588, A000009.
Cf. A001318, A193828.
Cf. A014493, A128880, A154293.

With[{nn=50},Sort[Select[Table[(n(3n-1))/2,{n,-nn,nn}],OddQ]]] (* Harvey P. Dale, Feb 16 2014 *)

Sum_{n>=1} 1/a(n) = Pi/2. - Amiram Eldar, Aug 18 2022

Corrected by T. D. Noe, Oct 25 2006

A069497 Triangular numbers of the form 6*k.

Original entry on oeis.org

0, 6, 36, 66, 78, 120, 210, 276, 300, 378, 528, 630, 666, 780, 990, 1128, 1176, 1326, 1596, 1770, 1830, 2016, 2346, 2556, 2628, 2850, 3240, 3486, 3570, 3828, 4278, 4560, 4656, 4950, 5460, 5778, 5886, 6216, 6786, 7140, 7260, 7626, 8256, 8646, 8778, 9180

Offset: 1

Amarnath Murthy, Mar 30 2002

Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Cf. A000217, A108752, A112652, A154293.

a[0] := 0:a[1] := 3:a[2] := 8:a[3] := 11:seq((12*(floor(i/4))+a[i mod 4])*(12*(floor(i/4))+a[i mod 4]+1)/2,i=0..100);

CoefficientList[ Series[ 6x (x^2 -x +1) (x^2 +4x +1)/((x^2 +1)^2*(1 -x)^3), {x, 0, 45}], x] (* or *)
LinearRecurrence[{3, -5, 7, -7, 5, -3, 1}, {0, 6, 36, 66, 78, 120, 210}, 46] (* Robert G. Wilson v, May 31 2017 *)
Select[Accumulate[Range[0, 89]], Divisible[#, 6] &] (* Alonso del Arte, May 31 2017 *)

a(n) = 6 * A154293(n). - Joerg Arndt, Aug 18 2022
a(n) = A000217(A112652(n+1)-1). - R. J. Mathar, Aug 21 2007
From R. J. Mathar, Nov 18 2009: (Start)
a(n) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7).
G.f.: 6*x*(x^2-x+1)*(x^2+4*x+1)/((1+x^2)^2*(1-x)^3) (6*A154293). (End)
From Amiram Eldar, Aug 18 2022: (Start)
a(n) = A000217(A108752(n)).
Sum_{n>=2} 1/a(n) = 2 - (3+4*sqrt(3))*Pi/18. (End)

More terms from Sascha Kurz, Apr 01 2002
More terms from R. J. Mathar, Aug 21 2007
Offset corrected to 1, Joerg Arndt, Aug 18 2022

A154303 Primes of the form (1+2+...+m)/90 = A000217(m)/90.

Original entry on oeis.org

7, 11, 179, 181

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=90."

The corresponding m-values are m=35,44,179,180. It is clear that for m>180, T(m)/90 = m(m+1)/180 cannot be a prime, since then each factor in the numerator is larger than the denominator. All of the sequences A154296, ..., A154304 could or should be grouped together in a single ("fuzzy"?) table. It would be more interesting to have the function f(n) which gives the *number* of primes of the form T(k)/n. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/90;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,5*9!}];lst

d=90*2;for(m=1,999,(m^2+m)%d==0&isprime((m^2+m)/d)&print1(m",")) \\ print the m-values(!) - use A154304(90) to get A154303 as a list/vector. \\ - M. F. Hasler, Jan 06 2013

Edited by M. F. Hasler, Jan 06 2013

A164577 Integer averages of the first perfect cubes up to some n^3.

Original entry on oeis.org

1, 12, 25, 45, 112, 162, 225, 396, 507, 637, 960, 1156, 1377, 1900, 2205, 2541, 3312, 3750, 4225, 5292, 5887, 6525, 7936, 8712, 9537, 11340, 12321, 13357, 15600, 16810, 18081, 20812, 22275, 23805, 27072, 28812, 30625, 34476, 36517, 38637, 43120

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Integers of the form A000537(k)/k, created by the k>0 listed in A042965. - R. J. Mathar, Aug 20 2009

Integers of the form (1/4)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009

			The average of the first cube is 1^3/1=1=a(1).
The average of the first two cubes is (1^3+2^3)/2=9/2, not integer, and does not contribute to the sequence.
The average of the first three cubes is (1^3+2^3+3^3)/3=12, integer, and defines a(2).

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

Cf. A000537, A050248, A051456, A078617, A078618, A154293, A164576

Timing[s=0;lst={};Do[a=(s+=n^3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n, 5!}];lst]
With[{nn=80},Select[#[[1]]/#[[2]]&/@Thread[{Accumulate[Range[ nn]^3],Range[ nn]}],IntegerQ]] (* or *) LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{1,12,25,45,112,162,225,396,507,637},50] (* Harvey P. Dale, Mar 14 2020 *)

G.f.: ( x*(1+11*x+13*x^2+17*x^3+34*x^4+11*x^5+6*x^6+3*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011

Changed comments to examples - R. J. Mathar, Aug 20 2009

A268539 Numbers k such that 48*k + 25 is a perfect square.

Original entry on oeis.org

0, 2, 3, 7, 17, 25, 28, 38, 58, 72, 77, 93, 123, 143, 150, 172, 212, 238, 247, 275, 325, 357, 368, 402, 462, 500, 513, 553, 623, 667, 682, 728, 808, 858, 875, 927, 1017, 1073, 1092, 1150, 1250, 1312, 1333, 1397, 1507, 1575, 1598, 1668, 1788, 1862, 1887, 1963, 2093, 2173

Offset: 1

N. J. A. Sloane, Feb 24 2016

Equivalently, integers of the form (h+5)*(h-5)/48, where h must be odd, h = 2*m+1, thus also integers of the form (m+3)*(m-2)/12, with m = 2, 5, 6, 9, 14, 17, 18, ... = {2, 5, 6, 9} + 12 N. - M. F. Hasler, Mar 02 2016
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(8*n))*(1 + q^(8*n-1))*(1 + q^(8*n-7))/(1 + q^n) = Sum_{n >= 0} q^(2*n*(n+1)) * Product_{k >= 2*n+2} 1 - q^k = 1 - q^2 - q^3 + q^7 + q^17 - q^25 - q^28 + + - - ... (by the quintuple product identity and Mc Laughlin et al., S.38, p 16). - Peter Bala, Dec 30 2024
Conjecture: the sequence terms are also the exponents in the expansion of Sum_{n >= 0} q^n/(Product_{k = 1..2*n+1} 1 + q^k) = 1 + q^2 - q^3 - q^7 + q^17 + q^25 - - + + .... - Peter Bala, Jan 15 2025

Zak Seidov, Table of n, a(n) for n = 1..10000
J. Mc Laughlin, A. V. Sills and P. Zimmer, Rogers-Ramanujan-Slater Type Identities, Electronic J. Combinatorics, DS15, 1-59, May 31, 2008.
Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157, December 2015, Pages 223-232, see Corollary 4.4.
Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Cf. A078405, A154293.

Subsequence of A011865.

[n: n in [0..2200] | IsSquare(48*n+25)]; // Vincenzo Librandi, Feb 25 2016

L := [5, 11, 13, 19, 29, 35, 37, 43]:
seq(seq(((L[i]+48*j)^2-25)/48, i=1..8), j=0..10); # Robert Israel, Feb 29 2016

Select[Range[0, 2500], IntegerQ[Sqrt[48 # + 25]] &] (* Vincenzo Librandi, Feb 25 2016 *)
Table[(3 (n - 1) n + (2 n - 1) (-1)^((n - 2) (n - 1)/2) - 1)/4, {n, 1, 60}] (* Bruno Berselli, Feb 29 2016 *)
LinearRecurrence[{3, -5, 7, -7, 5, -3, 1}, {0, 2, 3, 7, 17, 25, 28}, 48] (* Robert G. Wilson v, Mar 05 2016 *)
CoefficientList[ Series[ x*(2 - 3x + 8x^2 - 3x^3 + 2x^4)/((1 - x)^3*(1 + x^2)^2), {x, 0, 47}], x] (* Robert G. Wilson v, Mar 05 2016 *)

isok(n) = issquare(48*n+25); \\ Michel Marcus, Feb 25 2016

A268539(n)={my(m=n\4*12+[-3,2,5,6][n%4+1]);(3+m)*(m-2)/12} \\ M. F. Hasler, Mar 03 2016

from gmpy2 import is_square
[k for k in range(2200) if is_square(48*k+25)] # Bruno Berselli, Dec 05 2016

[n for n in (0..2200) if is_square(48*n+25)] # Bruno Berselli, Feb 29 2016

For n>25, a(n) = 3*( a(n-8)-a(n-16) ) + a(n-24). - Zak Seidov, Feb 28 2016
From Robert Israel, Feb 29 2016: (Start)
Let L = [5, 11, 13, 19, 29, 35, 37, 43].
Then a(i + 8*j) = ( (L(i) + 48*j)^2 - 25 )/48 for i = 1..8, j >= 0. (End)
From Bruno Berselli, Feb 29 2016: (Start)
G.f.: x^2*(2 - 3*x + 8*x^2 - 3*x^3 + 2*x^4)/((1 - x)^3*(1 + x^2)^2).
a(n) = a(-n+1) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7) for n>6.
a(n) = (3*(n-1)*n + (2*n-1)*(-1)^((n-2)*(n-1)/2) - 1)/4. Therefore:
a(4*k)   = k*(12*k -5),
a(4*k+1) = k*(12*k +5),
a(4*k+2) = k*(12*k+11)+2 = (3*k+2)*(4*k+1),
a(4*k+3) = k*(12*k+13)+3 = (3*k+1)*(4*k+3).
From the previous formulas follows that 2, 3, 7 and 17 are the only primes of the sequence. (End)
Sum_{n>=2} 1/a(n) = 12/25 + (4/sqrt(3)-1)*Pi/5. - Amiram Eldar, Jul 30 2024

More terms from Michel Marcus, Feb 25 2016

cp's OEIS Frontend

A287746 First differences of A154293.

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A057570 Numbers of the form n*(7n+-1)/2.

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A154296 Primes of the form (1+2+3+...+m)/15 = A000217(m)/15, for some m.

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A154304 Primes of the form (1+2+...+m)/210 = A000217(m)/210.

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A164576 Integer averages of the set of the first positive squares up to some n^2.

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A067589 Numbers k such that A067588(k) is an odd number.

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A069497 Triangular numbers of the form 6*k.

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A154303 Primes of the form (1+2+...+m)/90 = A000217(m)/90.

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