A139282 -id:A139282 - cp's OEIS frontend

A139275 a(n) = n(8n+1).

0, 9, 34, 75, 132, 205, 294, 399, 520, 657, 810, 979, 1164, 1365, 1582, 1815, 2064, 2329, 2610, 2907, 3220, 3549, 3894, 4255, 4632, 5025, 5434, 5859, 6300, 6757, 7230, 7719, 8224, 8745, 9282, 9835, 10404, 10989, 11590, 12207, 12840

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 9,..., in the square spiral whose vertices are the triangular numbers A000217.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 1), {n, 0, 40}] (* Bruno Berselli, Sep 21 2016 *)
LinearRecurrence[{3,-3,1},{0,9,34},50] (* Harvey P. Dale, Apr 21 2020 *)

a(n) = n*(8*n+1); \\ Altug Alkan, Sep 21 2016

a(n) = 8*n^2 + n.
Sequences of the form a(n) = 8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 7 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
a(n) = A000217(5*n) - A000217(3*n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} 1/a(n) = 8 - (1+sqrt(2))*Pi/2 - 4*log(2) - sqrt(2) * log(1+sqrt(2)) = 0.1887230016056779928... . - Vaclav Kotesovec, Sep 21 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(7*x + 9)/(1-x)^3.
E.g.f.: (8*x^2 + 9*x)*exp(x). (End)

A139271 a(n) = 2n(4*n-3).

Original entry on oeis.org

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.
Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Table[8n^2-6n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,2,20},50] (* Harvey P. Dale, Sep 26 2016 *)

a(n)=2*n*(4*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 - 6*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Corrected by Harvey P. Dale, Sep 26 2016

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A139278 a(n) = n(8n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., in the square spiral whose vertices are the triangular numbers A000217.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139276, A139277, A139279, A139280, A139281, A139282.

Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)

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

a(n) = 8*n^2 + 7*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022

A139272 a(n) = n(8n-5).

Original entry on oeis.org

0, 3, 22, 57, 108, 175, 258, 357, 472, 603, 750, 913, 1092, 1287, 1498, 1725, 1968, 2227, 2502, 2793, 3100, 3423, 3762, 4117, 4488, 4875, 5278, 5697, 6132, 6583, 7050, 7533, 8032, 8547, 9078, 9625, 10188, 10767, 11362, 11973, 12600

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 3, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139276 in the same spiral.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(n*k-k+6)/2, this sequence is the case k=16: see Comments lines of A226492.

s=0;lst={s};Do[s+=n++ +3;AppendTo[lst, s], {n, 0, 7!, 16}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
Table[n*(8*n -5), {n,0,50}] (* G. C. Greubel, Jul 18 2017 *)
LinearRecurrence[{3,-3,1},{0,3,22},50] (* Harvey P. Dale, Jan 13 2024 *)

a(n)=n*(8*n-5) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 8*n^2 - 5*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 13 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(13*x + 3)/(1-x)^3.
E.g.f.: (8*x^2 + 3*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = ((sqrt(2)-1)*Pi + 8*log(2) - 2*sqrt(2)*log(sqrt(2)+1))/10. - Amiram Eldar, Mar 17 2022

A139274 a(n) = n(8n-1).

Original entry on oeis.org

0, 7, 30, 69, 124, 195, 282, 385, 504, 639, 790, 957, 1140, 1339, 1554, 1785, 2032, 2295, 2574, 2869, 3180, 3507, 3850, 4209, 4584, 4975, 5382, 5805, 6244, 6699, 7170, 7657, 8160, 8679, 9214, 9765, 10332, 10915, 11514, 12129, 12760, 13407, 14070, 14749

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the triangular numbers A000217.

Polygonal number connection: 2*P_n + 5*S_n where P_n is the n-th pentagonal number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010

			a(1) = 16*1 + 0 - 9 = 7; a(2) = 16*2 + 7 - 9 = 30; a(3) = 16*3 + 30 - 9 = 69. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

[n*(8*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

A139274:=n->n*(8*n-1): seq(A139274(n), n=0..100); # Wesley Ivan Hurt, Dec 04 2016

CoefficientList[Series[x (9 x + 7)/(1 - x)^3, {x, 0, 43}], x] (* Michael De Vlieger, Jan 11 2020 *)
Table[n(8n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,30},50] (* Harvey P. Dale, Apr 01 2024 *)

a(n)=n*(8*n-1) \\ Charles R Greathouse IV, Jun 17 2017

Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = (1/3) * Sum_{i=n..(7*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(9*x+7)/(1-x)^3.
E.g.f.: (8*x^2 + 7*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 4*log(2) + sqrt(2)*log(sqrt(2)+1) - (sqrt(2)+1)*Pi/2. - Amiram Eldar, Mar 18 2022

A139276 a(n) = n(8n+3).

Original entry on oeis.org

0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.

			a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139277, A139278, A139279, A139280, A139281, A139282.

a[n_]:=n*(8*n+3); a[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3,-3,1},{0,11,38},50] (* Harvey P. Dale, Jul 02 2022 *)

a(n)=n*(8*n+3) \\ Charles R Greathouse IV, Jun 16 2017

a(n) = 8*n^2 + 3*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022

A139277 a(n) = n(8n+5).

Original entry on oeis.org

0, 13, 42, 87, 148, 225, 318, 427, 552, 693, 850, 1023, 1212, 1417, 1638, 1875, 2128, 2397, 2682, 2983, 3300, 3633, 3982, 4347, 4728, 5125, 5538, 5967, 6412, 6873, 7350, 7843, 8352, 8877, 9418, 9975, 10548, 11137, 11742, 12363, 13000

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139273 in the same spiral.

Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250.

Cf. A139271, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 5), {n, 0, 50}] (* Bruno Berselli, Aug 22 2018 *)
LinearRecurrence[{3,-3,1},{0,13,42},50] (* Harvey P. Dale, Dec 04 2018 *)

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

a(n) = 8*n^2 + 5*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x*{c+8 + (8-c)*x}/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 3 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
Sum_{n>=1} 1/a(n) = (sqrt(2)-1)*Pi/10 - 4*log(2)/5 + sqrt(2)*log(sqrt(2)+1)/5 + 8/25. - Amiram Eldar, Mar 18 2022
E.g.f.: exp(x)*x*(13 + 8*x). - Elmo R. Oliveira, Dec 15 2024

A329447 Start with a(0)=0; thereafter, look left and identify the least frequent digit d so far (in case of a tie, choose the smallest d): after then a(n) = 10c + d, where c > 0 is the number of times d has appeared so far.

Original entry on oeis.org

0, 10, 11, 20, 12, 22, 30, 13, 23, 33, 40, 14, 24, 34, 44, 50, 15, 25, 35, 45, 55, 60, 16, 26, 36, 46, 56, 66, 70, 17, 27, 37, 47, 57, 67, 77, 80, 18, 28, 38, 48, 58, 68, 78, 88, 90, 19, 29, 39, 49, 59, 69, 79, 89, 99, 100, 112, 113, 114, 115, 116, 117, 118, 119, 120, 123, 124, 125, 126, 127, 128, 129

Offset: 0

Eric Angelini and M. F. Hasler, Nov 14 2019

The term "10c + d" are to be read "c digits d have appeared so far", as in the "look and see" sequences llike A045918.
It follows immediately from the definition that all terms are distinct. For a sorted list of the terms, see A376779. For a tabular method of computing a(n), see the triangle in A377905. - N. J. A. Sloane, Nov 11 2024
An analogous sequence may be obtained for any initial term a(0). Sequence A329448 lists the starting values that will appear a second time later in the respective sequence.

M. F. Hasler, Table of n, a(n) for n = 0..10000 (independently computed by _Jean-Marc Falcoz_).
Eric Angelini, Look left and say, Cinquante signes Blog Post, Nov 14 2019.
Eric Angelini, Look left and say, Cinquante signes Blog Post, Nov 14 2019. [Local copy, with permission]

Cf. A045918 (describe n), A005150 (the classic "Say What You See"), A139282 (count vowels so far), A139097 (count letters so far).

A376779 gives terms in increasing order. See also A377905.

a[0]:= 0;
S[0]:= 1:
for i from 1 to 9 do S[i]:= 0 od:
for n from 1 to 100 do
  a[n]:= min(select(`>=`,[seq(10*S[i]+i, i=0..9)],10));
  L:= convert(a[n],base,10);
  for d from 0 to 9 do S[d]:= S[d] + numboccur(d,L) od;
od:
seq(a[n],n=0..100); # Robert Israel, Nov 14 2019

A329447_vec(N)={my(c=Vec(1,10),t); vector(N,i, for(j=1, #i=vecsort(c,,1), if(c[i[j]], i=i[j];break)); for(j=1, #i=digits(t=c[i]*10+i-1), c[i[j]+1]++);t)} \\ Returns the vector a(1..N)

from itertools import islice
def agen():  # generator of terms
    counts = [1] + [0 for i in range(1, 10)]
    yield 0
    while True:
        m = float('inf')
        for i in range(10):
            if counts[i] and counts[i] < m:
                m, argm = counts[i], i
        an = 10*m + argm
        yield an
        for d in str(an): counts[int(d)] += 1
print(list(islice(agen(), 72))) # Michael S. Branicky, Nov 11 2024

Edited by N. J. A. Sloane, Nov 10 2024

A139212 Number of vowels in the preceding terms spelled out in French.

Original entry on oeis.org

0, 2, 4, 7, 8, 10, 11, 13, 16, 19, 22, 25, 27, 29, 32, 36, 39, 43, 49, 55, 60, 64, 71, 78, 85, 90, 95, 102, 105, 107, 109, 112, 116, 120, 122, 126, 129, 133, 138, 143, 150, 155, 161, 168, 175, 183, 190, 196, 204, 210, 214, 221, 227, 232, 239, 246, 254, 264, 274, 285, 293, 303, 308, 313, 319, 325, 330, 335, 341, 350, 357, 365, 373, 383

Offset: 1

N. J. A. Sloane (based on Angelini's article), Jun 08 2008

Form a sequence of French words as follows: look to the left, towards the beginning of the sequence and write down the number of vowels you see; repeat; then replace the words by the corresponding numbers.
The sequence of words is: zero, deux, quatre, sept, huit, dix, onze, treize, seize, ...
Hyphens, accents and spaces are not counted.

			The second word is "deux" (and so a(2)=2 for the 'e' and the 'u'), because at the end of the first word ("zéro") we can see two vowels ('e' and 'o') to the left.

E. Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.

For an English version see A139282.

a(n)={ n>1 || return;#select(Vec(French(n=a(n-1))),x->setsearch(Vec("aeiou"),x))+n }  /* see A167507 for French() */
/* Version with memoization for better performance when n >> 100: */
A139212(n)={ type(a139212)!="t_VEC" && a139212=[];
n > #a139212 && a139212=concat( a139212,vector(n-#a139212));
(a139212[n] || n==1) && return(a139212[n]);
a139212[n]=/*up to here only memoization, could be omitted*/
#select(Vec(French(A139212(n-1))), x->setsearch(Vec("aeiou"),x))) + A139212(n-1) } \\ M. F. Hasler, Sep 29 2011

Fixed offset (according to example) and typo in example. M. F. Hasler, Sep 29 2011

cp's OEIS Frontend

A139275 a(n) = n*(8*n+1).

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A139271 a(n) = 2*n*(4*n-3).

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A139273 a(n) = n*(8*n - 3).

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Magma

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A139278 a(n) = n*(8*n+7).

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A139272 a(n) = n*(8*n-5).

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A139274 a(n) = n*(8*n-1).

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A139276 a(n) = n*(8*n+3).

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A139275 a(n) = n(8n+1).

A139271 a(n) = 2n(4*n-3).

A139273 a(n) = n(8n - 3).

A139278 a(n) = n(8n+7).

A139272 a(n) = n(8n-5).

A139274 a(n) = n(8n-1).

A139276 a(n) = n(8n+3).

A139277 a(n) = n(8n+5).