Raphael Ranna - cp's OEIS frontend

A343007 Relative position of the average value between two consecutive partial sums of the Leibniz formula for Pi.

6, 13, 26, 41, 62, 85, 114, 145, 182, 221, 266, 313, 366, 421, 482, 545, 614, 685, 762, 841, 926, 1013, 1106, 1201, 1302, 1405, 1514, 1625, 1742, 1861, 1986, 2113, 2246, 2381, 2522, 2665, 2814, 2965, 3122, 3281, 3446, 3613, 3786, 3961, 4142, 4325, 4514, 4705

Offset: 1

Raphael Ranna, Apr 02 2021

Define L(n) to be the n-th partial sum of the Leibniz formula Pi = 4 - 4/3 + 4/5 - 4/7 + ..., i.e., L(n) = Sum_{j=1..n} 4*(-1)^(j+1)/(2*j-1). For every positive integer n, L(n+1) is closer to Pi than L(n) is. If we let V be the average of the two consecutive partial sums L(n) and L(n+1), then the partial sums that lie closest to V are L(a(n)-1) and L(a(n)+1) (one of which is above V, the other below).

			The first several partial sums are as follows:
  n      L(n)
  -  ------------
  1  4.0000000000
  2  2.6666666...
  3  3.4666666...
  4  2.8952380...
  5  3.3396825...
  6  2.9760461...
  7  3.2837384...
  8  3.0170718...
.
For n=1, the average of the partial sums L(1) and L(2) is V = (L(1) + L(2))/2 = (4 + 2.6666666...)/2 = 3.3333333...; the two partial sums closest to V are L(5)=3.3396825... and L(7)=3.2837384..., and V lies in the interval between them, so a(1)=6.
The formula as it is written works for all data in the sequence, but it needs to be proven that it works for all possible integer values of n.

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A047550, A102083, A254527.

Rest@ CoefficientList[Series[x (6 + x + x^3)/((1 + x) (1 - x)^3), {x, 0, 48}], x] (* Michael De Vlieger, Apr 05 2021 *)

a(1) = 6; a(n) = a(n-1) + r(n), where r(n) = A047550(n) = 4*n - (-1)^n.
G.f.: x*(6 + x + x^3)/((1 + x)*(1 - x)^3). - Jinyuan Wang, Apr 03 2021
From Stefano Spezia, Apr 03 2021: (Start)
a(n) = (3 + (-1)^(n+1) + 4*n + 4*n^2)/2.
a(2*n) = A102083(n).
a(2*n-1) = A254527(n). (End)

A275464 a(n) = n - A038802(n).

Original entry on oeis.org

0, 0, 0, 3, 1, 1, 6, 2, 2, 9, 3, 10, 12, 5, 5, 15, 15, 7, 18, 8, 8, 21, 9, 21, 24, 11, 25, 27, 13, 13, 30, 30, 15, 33, 16, 16, 36, 35, 18, 39, 19, 40, 42, 21, 42, 45, 45, 24, 48, 25, 25, 51, 26, 26, 54, 27, 55, 57, 56, 56, 60, 60, 33, 63, 34, 63, 66, 36, 36, 69, 67, 70, 72, 40, 40, 75, 75, 42, 78, 77, 44, 81

Offset: 1

Raphael Ranna, Jul 28 2016

nonn

			9 = (2^0)*(3^2), resulting in 1 zero. So, a(4) = 4 - 1 = 3.

Raphael Ranna, Table of n, a(n) for n = 1..1000

Cf. A038802.

Table[f = FactorInteger[2 n + 1]; n - (PrimePi[f[[1, 1]]] - 1), {n, 100}]

lpf(n)=factor(n)[1,1]
a(n)=n - primepi(lpf(2*n+1)) + 1 \\ Charles R Greathouse IV, Jul 29 2016

a(n) = n - (A049084(A020639(2n+1)) - 1).

a(n) = n + O(n/log n). - Charles R Greathouse IV, Jul 29 2016

Duplicate a(44)-a(45) removed by Andrew Howroyd, Feb 23 2018

A261574 a(n) = n(n^2 + 3)(n^6 + 6n^4 + 9n^2 + 3).

Original entry on oeis.org

0, 76, 2786, 46764, 439204, 2744420, 12813606, 48229636, 153992264, 432083484, 1092730090, 2537720636, 5489037036, 11179326964, 21624372014, 40001698260, 71163830416, 122319408236, 203920464114, 330799604044, 523606640180, 810600392196, 1229857906486

Offset: 0

Raphael Ranna, Aug 24 2015

Also numbers of the form (n-th metallic mean)^9 - 1/(n-th metallic mean)^9, see link to Wikipedia.

Colin Barker, Table of n, a(n) for n = 0..1000 (first 100 terms from Raphael Ranna)
Wikipedia, Metallic mean
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A001622, A014176, A098316, A098317, A098318, A176398, A176439, A176458, A176522, A261391, A261540.

[n*(n^2+3)*(n^6+6*n^4+9*n^2+3): n in [0..25]]; // Bruno Berselli, Aug 25 2015

Table[n (n^2 + 3) (n^6 + 6 n^4 + 9 n^2 + 3), {n, 0, 25}] (* Bruno Berselli, Aug 25 2015 *)

concat(0, Vec(2*x*(38*x^8 +1013*x^7 +11162*x^6 +43907*x^5 +69200*x^4 +43907*x^3 +11162*x^2 +1013*x +38) / (x -1)^10 + O(x^50))) \\ Colin Barker, Aug 25 2015

a(n) = -a(-n) = ( (n+sqrt(n^2+4))/2 )^9-1/( (n+sqrt(n^2+4))/2 )^9.

G.f.: 2*x*(38*x^8 +1013*x^7 +11162*x^6 +43907*x^5 +69200*x^4 +43907*x^3 +11162*x^2 +1013*x +38) / (x -1)^10. - Colin Barker, Aug 25 2015

Formula in Name by Colin Barker, Aug 25 2015

Offset changed from 1 to 0 and initial 0 added by Bruno Berselli, Aug 25 2015

A261540 a(n) = n^7 + 7n^5 + 14n^3 + 7*n.

Original entry on oeis.org

0, 29, 478, 4287, 24476, 101785, 337434, 946043, 2333752, 5206581, 10714070, 20633239, 37597908, 65378417, 109216786, 176222355, 275832944, 420346573, 625528782, 911300591, 1302512140, 1829807049, 2530582538, 3450050347, 4642403496, 6172093925, 8115226054

Offset: 0

Raphael Ranna, Aug 24 2015

Also numbers of the form (n-th metallic mean)^7 - 1/(n-th metallic mean)^7, see link to Wikipedia.

Raphael Ranna, Table of n, a(n) for n = 0..100
Wikipedia, Metallic mean
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A001622, A014176, A098316, A098317, A098318, A176398, A176439, A176458, A176522, A261391.

[n^7 + 7*n^5 + 14*n^3 + 7*n: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

Table[n^7 + 7 n^5 + 14 n^3 + 7 n, {n, 0, 30}] (* Bruno Berselli, Aug 24 2015 *)
LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 29, 478, 4287, 24476, 101785, 337434, 946043}, 30] (* Vincenzo Librandi, Aug 24 2015 *)

a(n)=n^7+7*n^5+14*n^3+7*n \\ Charles R Greathouse IV, Aug 24 2015

[n^7+7*n^5+14*n^3+7*n for n in (0..30)] # Bruno Berselli, Aug 24 2015

a(n) = -a(-n) = ( (n+sqrt(n^2+4))/2 )^7 - 1/( (n+sqrt(n^2+4))/2 )^7.

G.f.: x*(29 + 246*x + 1275*x^2 + 1940*x^3 + 1275*x^4 + 246*x^5 + 29*x^6)/(1 - x)^8. - Bruno Berselli, Aug 24 2015

Offset changed from 1 to 0 and initial 0 added by Bruno Berselli, Aug 25 2015

A261391 a(n) = n^5 + 5n^3 + 5n.

Original entry on oeis.org

0, 11, 82, 393, 1364, 3775, 8886, 18557, 35368, 62739, 105050, 167761, 257532, 382343, 551614, 776325, 1069136, 1444507, 1918818, 2510489, 3240100, 4130511, 5206982, 6497293, 8031864, 9843875, 11969386, 14447457, 17320268, 20633239, 24435150, 28778261, 33718432, 39315243, 45632114

Offset: 0

Raphael Ranna, Aug 17 2015

Also numbers of the form (n-th metallic mean)^5 - 1/(n-th metallic mean)^5, see link to Wikipedia.

Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, Metallic mean.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A001622, A014176, A098316, A098317, A098318, A176398, A176439, A176458, A176522.

Array[#^5 + 5 #^3 + 5 # &, 34] (* Michael De Vlieger, Aug 18 2015 *)
Table[n^5 + 5*n^3 + 5*n, {n,0, 50}] (* G. C. Greubel, Aug 21 2015 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,11,82,393,1364,3775},40] (* Harvey P. Dale, May 07 2018 *)

concat(0, Vec(x*(11*x^4+16*x^3+66*x^2+16*x+11)/(x-1)^6 + O(x^100))) \\ Colin Barker, Aug 18 2015

a(n) = ( (n+sqrt(n^2+4))/2 )^5 - 1/( (n+sqrt(n^2+4))/2 )^5.
a(n) = -a(-n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Colin Barker, Aug 18 2015
G.f.: x*(11*x^4+16*x^3+66*x^2+16*x+11) / (x-1)^6. - Colin Barker, Aug 18 2015
E.g.f.: (x^5 + 15*x^4 + 70*x^3 + 120*x^2 + 71*x + 11)*e^x. - G. C. Greubel, Aug 21 2015

Offset changed from 1 to 0, initial 0 added and b-file adapted from Bruno Berselli, Aug 25 2015

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User: Raphael Ranna

A343007 Relative position of the average value between two consecutive partial sums of the Leibniz formula for Pi.

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A275464 a(n) = n - A038802(n).

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A261574 a(n) = n*(n^2 + 3)*(n^6 + 6*n^4 + 9*n^2 + 3).

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A261540 a(n) = n^7 + 7*n^5 + 14*n^3 + 7*n.

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A261391 a(n) = n^5 + 5*n^3 + 5*n.

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A261574 a(n) = n(n^2 + 3)(n^6 + 6n^4 + 9n^2 + 3).

A261540 a(n) = n^7 + 7n^5 + 14n^3 + 7*n.

A261391 a(n) = n^5 + 5n^3 + 5n.