Joseph Myers - cp's OEIS frontend

A279766 Number of odd digits in the decimal expansions of integers 1 to n.

0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 40, 40, 41, 41, 42, 42, 43, 43, 44, 44, 45, 46, 48, 49, 51, 52, 54, 55, 57, 58, 60, 60, 61, 61, 62, 62, 63, 63, 64, 64

Offset: 0

Joseph Myers, Dec 18 2016

From Bernard Schott, Feb 19 2023: (Start)
Problem 1 of the British Mathematical Olympiad, round 1, in 2016/2017 asked: when the integers 1, 2, 3, ..., 2016 are written down in base 10, how many of the digits in the list are odd? The answer is a(2016) = 4015.
The similar sequence but with number of even digits is A358854. (End)

Joseph Myers, Table of n, a(n) for n = 0..1000
United Kingdom Mathematics Trust, 2016/17 British Mathematical Olympiad Round 1, Problem 1.
Index to sequences related to Olympiads.

Cf. A007908, A058183, A117804, A196564, A358854.

a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+
      nops(select(x-> x::odd, convert(n,base,10))))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Dec 22 2016

Table[Count[Flatten@ IntegerDigits@ Range[0, n], d_ /; OddQ@ d], {n, 0, 68}] (* or *)
Accumulate@ Table[Count[IntegerDigits@ n, d_ /; OddQ@ d], {n, 0, 68}] (* Michael De Vlieger, Dec 22 2016 *)

a(n) = A196564(A007908(n)). - Michel Marcus, Dec 18 2016

a(n) = A117804(n+1) - A358854(n) (number of total digits - number of even digits). - Bernard Schott, Feb 19 2023

A279259 Smallest positive integer m such that m, m+1, m+2, m+3 are divisible by 2n+1, 2n+3, 2n+5, 2n+7 respectively.

Original entry on oeis.org

53, 159, 1735, 4508, 3222, 18238, 31499, 16965, 78013, 114722, 54348, 225124, 303425, 133515, 519187, 662408, 277794, 1035370, 1272023, 515697, 1864393, 2228174, 880920, 3112528, 3642317, 1412343, 4901599, 5641460, 2154030, 7368982, 8368163, 3155229, 10667605, 11980538

Offset: 0

Joseph Myers, Dec 08 2016

			53 is the smallest positive integer such that 53, 54, 55, 56 are divisible by 1, 3, 5, 7 respectively, hence a(0) = 53. - _Bernard Schott_, Dec 08 2020

Joseph Myers, Table of n, a(n) for n = 0..1000
United Kingdom Mathematics Trust, 2016/17 British Mathematical Olympiad Round 1, Problem 6.
Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-10,0,0,10,0,0,-5,0,0,1).
Index to sequences related to Olympiads and other Mathematical competitions.

LinearRecurrence[{0,0,5,0,0,-10,0,0,10,0,0,-5,0,0,1},{53,159,1735,4508,3222,18238,31499,16965,78013,114722,54348,225124,303425,133515,519187},40] (* Harvey P. Dale, Dec 29 2017 *)

a(n) = (2*n+1 + lcm(2*n+1, 2*n+3, 2*n+5, 2*n+7))/2.
G.f.: (8*x^14 +10*x^12 -89*x^11 -153*x^10 -1777*x^9 -4173*x^8 -2445*x^7 -9489*x^6 -9563*x^5 -2427*x^4 -4243*x^3 -1735*x^2 -159*x-53) / ((x-1)^5*(x^2+x+1)^5). - Alois P. Heinz, Dec 08 2016
From Bernard Schott, Dec 08 2020: (Start)
If n == 1 (mod 3), a(n) = (2*n+1)* ((2*n+3)*(2*n+5)*(2*n+7)/3 + 1)/2.
If n == 0, 2 (mod 3), a(n) = (2*n+1)* ((2*n+3)*(2*n+5)*(2*n+7) + 1)/2. (End)

A277608 Least number of fractions of the form (k+1)/k, for k a positive integer, whose product equals n.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 6, 7, 6, 7, 5, 6, 6, 7, 6, 7, 7, 7, 6, 7, 7, 7, 7, 7, 8, 8, 6, 7, 7, 7, 7, 8, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 8, 8, 8, 9, 7, 8, 8, 9, 8, 8, 8, 9, 8

Offset: 1

Joseph Myers, Oct 23 2016

nonn

If each intermediate product of the first j of the fractions, for all j < a(n), is also restricted to be an integer, the resulting sequence is A117497. The first n for which a shorter product can be obtained by allowing intermediate non-integer products is 43 = 2/1 * 2/1 * 2/1 * 2/1 * 2/1 * 4/3 * 129/128, a product of 7 fractions, where A117497(43) = 8.

Joseph Myers, Table of n, a(n) for n = 1..10000
United Kingdom Mathematics Trust, Mathematical Olympiad for Girls 2016, problem 5.

Cf. A117497 (restriction to intermediate products being integers), A014701 (always generating n from n-1 for n odd and from n/2 for n even), A376012.

A264998 Number of partitions of n into distinct parts of the form 3^a*5^b or 2.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 3, 2, 3, 3, 2, 3, 2, 1, 2, 2, 3, 3, 4, 4, 4, 6, 4, 5, 5, 4, 5, 4, 4, 3, 4, 4, 4, 6, 5, 5, 7, 5, 5, 6, 4, 6, 6, 6, 6, 7, 7, 6, 8, 5, 6, 7, 5, 6, 5, 4, 4, 4, 4, 4, 5, 6, 5, 7, 6, 5, 9, 7, 8, 9, 7, 8, 9, 8, 7, 10, 8, 9, 11

Offset: 0

Joseph Myers, Nov 29 2015

			15 = 15 = 9 + 5 + 1 = 9 + 3 + 2 + 1, so a(15) = 3.

Joseph Myers and Reinhard Zumkeller, Table of n, a(n) for n = 0..20000 (first 1000 terms from Joseph Myers)
British Mathematical Olympiad 2015/16, Olympiad Round 1, Problem 6, Friday, 27 November 2015.

Cf. A003593, A264997.

import Data.MemoCombinators (memo2, list, integral)
a264998 n = a264998_list !! (n-1)
a264998_list = f 0 [] (1 : 2 : tail a003593_list) where
   f u vs ws'@(w:ws) | u < w = (p' vs u) : f (u + 1) vs ws'
                     | otherwise = f u (vs ++ [w]) ws
   p' = memo2 (list integral) integral p
   p _  0 = 1
   p [] _ = 0
   p (k:ks) m = if m < k then 0 else p' ks (m - k) + p' ks m
-- Reinhard Zumkeller, Dec 18 2015

nmax = 100; A003593 = Select[Range[nmax], PowerMod[15, #, #] == 0 &]; CoefficientList[Series[(1 + x^2) * Product[(1 + x^(A003593[[k]])), {k, 1, Length[A003593]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 18 2015 *)

G.f.: (1+x)(1+x^2)(1+x^3)(1+x^5)(1+x^9)(1+x^15)....

A264997 Number of partitions of n into distinct parts of the form 3^a*5^b.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 0, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 0, 1, 2, 2, 1, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2, 1, 2, 3, 2, 3, 3, 2, 4, 3, 1, 3, 3, 3, 3, 3, 3, 4, 4, 2, 4, 3, 2, 4, 3, 2, 2, 2, 2, 2, 2, 2, 3, 4, 2, 3, 4, 2, 5, 5, 3, 4, 4, 4, 5, 4, 2, 6, 6, 3, 5

Offset: 0

Joseph Myers, Nov 29 2015

			28 = 27 + 1 = 25 + 3 = 15 + 9 + 3 + 1, so a(28) = 3.

Alois P. Heinz, Table of n, a(n) for n = 0..20000 (first 1001 terms from Joseph Myers)
British Mathematical Olympiad 2015/16, Olympiad Round 1, Problem 6, Friday, 27 November 2015.

Cf. A003593, A264998.

import Data.MemoCombinators (memo2, list, integral)
a264997 n = a264997_list !! (n-1)
a264997_list = f 0 [] a003593_list where
   f u vs ws'@(w:ws) | u < w = (p' vs u) : f (u + 1) vs ws'
                     | otherwise = f u (vs ++ [w]) ws
   p' = memo2 (list integral) integral p
   p _  0 = 1
   p [] _ = 0
   p (k:ks) m = if m < k then 0 else p' ks (m - k) + p' ks m
-- Reinhard Zumkeller, Dec 18 2015

nmax = 100; A003593 = Select[Range[nmax], PowerMod[15, #, #] == 0 &]; CoefficientList[Series[Product[(1 + x^(A003593[[k]])), {k, 1, Length[A003593]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 01 2015 *)

G.f.: (1+x)(1+x^3)(1+x^5)(1+x^9)(1+x^15)....

A227715 Triangle read by rows: Number of 2n-step self-avoiding walks on diamond lattice ending at point with x = 2k.

Original entry on oeis.org

1, 4, 4, 28, 24, 16, 188, 188, 128, 64, 1428, 1368, 1120, 640, 256, 10708, 10572, 8864, 6208, 3072, 1024, 82948, 81376, 71572, 53376, 32768, 14336, 4096, 644788, 637148, 570512, 453424, 304640, 166912, 65536, 16384, 5067404, 5007560, 4572076, 3762672, 2728256, 1669120

Offset: 0

Joseph Myers, Jul 21 2013

The number of walks ending with x = -k is the same as the number ending with x = k.

			Initial rows (paths of length 0, 2, 4, ...):
{ 1 };
{ 4, 4 };
{ 28, 24, 16 };
{ 188, 188, 128, 64 }.

Bert Dobbelaere, Table of n, a(n) for n = 0..135 (terms 0..77 from Joseph Myers)
J. L. Martin, The exact enumeration of self-avoiding walks on a lattice, Proc. Camb. Phil. Soc., 58 (1962), 92-101.

Cf. A001394, A001395, A001396, A001397, A001398, A227716.

A227716 Triangle read by rows: Number of (2n+1)-step self-avoiding walks on diamond lattice ending at point with x = 2k+1.

Original entry on oeis.org

2, 10, 8, 74, 56, 32, 518, 464, 288, 128, 3934, 3520, 2656, 1408, 512, 29914, 27768, 21920, 14336, 6656, 2048, 232094, 217316, 181456, 128256, 74240, 30720, 8192, 1812890, 1719616, 1475172, 1118592, 716288, 372736, 139264, 32768, 14277886, 13633972, 11989800, 9480048

Offset: 0

Joseph Myers, Jul 21 2013

The number of walks ending with x = -k is the same as the number ending with x = k.

			Initial rows (paths of length 1, 3, 5, ...):
{ 2 };
{ 10, 8 };
{ 74, 56, 32 };
{ 518, 464, 288, 128 }.

Bert Dobbelaere, Table of n, a(n) for n = 0..135 (terms 0..77 from Joseph Myers)
J. L. Martin, The exact enumeration of self-avoiding walks on a lattice, Proc. Camb. Phil. Soc., 58 (1962), 92-101.

Cf. A001394, A001395, A001396, A001397, A001398, A227715.

A227511 Triangle read by rows: Number of n-step self-avoiding walks on f.c.c. lattice ending at point with x = k.

Original entry on oeis.org

1, 4, 4, 36, 32, 16, 308, 292, 192, 64, 2764, 2672, 2016, 1024, 256, 25404, 24780, 20160, 12480, 5120, 1024, 237164, 232512, 197940, 137472, 71680, 24576, 4096, 2237948, 2201948, 1930944, 1443616, 869376, 390144, 114688, 16384, 21286548, 20997008, 18805488, 14786176

Offset: 0

Joseph Myers, Jul 14 2013

The number of walks ending with x = -k is the same as the number ending with x = k.

			Initial rows (paths of length 0, 1, 2, ...):
{ 1 };
{ 4, 4 };
{ 36, 32, 16 };
{ 308, 292, 192, 64 }.

Bert Dobbelaere, Table of n, a(n) for n = 0..135 (terms 0..77 from Joseph Myers)
J. L. Martin, The exact enumeration of self-avoiding walks on a lattice, Proc. Camb. Phil. Soc., 58 (1962), 92-101.

Cf. A000765, A000766, A000767, A000768, A001336.

A227338 Number of n-step self-avoiding walks on cubic lattice ending at point with x = k.

Original entry on oeis.org

1, 4, 1, 12, 8, 1, 44, 40, 12, 1, 172, 176, 84, 16, 1, 772, 748, 468, 144, 20, 1, 3308, 3248, 2332, 984, 220, 24, 1, 14924, 14280, 11068, 5756, 1788, 312, 28, 1, 64956, 63768, 51472, 30760, 12108, 2944, 420, 32, 1, 294252, 285296, 237832, 155912, 72948, 22732, 4516

Offset: 0

Joseph Myers, Jul 07 2013

The number of walks ending with x = -k is the same as the number ending with x = k.

			Initial rows (paths of length 0, 1, 2, ...):
1;
4, 1;
12, 8, 1;
44, 40, 12, 1;
...

Bert Dobbelaere, Table of n, a(n) for n = 0..275 (terms 0..152 from Joseph Myers)
J. L. Martin, The exact enumeration of self-avoiding walks on a lattice, Proc. Camb. Phil. Soc., 58 (1962), 92-101.

Cf. A000759, A000760, A000761, A000762, A001412.

For n > 0, A001412(n) = T(n,0) + 2 * Sum_{k=1..n} T(n,k). - Bert Dobbelaere, Jan 06 2019

A220350 Number of ways of putting 2n^2 counters on distinct squares of a 2n x 2n board so that no row, column or main diagonal contains more than n counters.

Original entry on oeis.org

0, 33, 112160, 43346734009, 2441203940002594824

Offset: 1

Joseph Myers, Dec 11 2012

			Configurations for n=2 (4x4 board) (o = counter):
.
oo.. (2 orientations)
o.o.
.o.o
..oo
.
.o.o (2 orientations)
oo..
..oo
o.o.
.
oo.. (4 orientations)
o..o
..oo
.oo.
.
oo.. (4 orientations)
..oo
oo..
..oo
.
oo.. (4 orientations)
..oo
..oo
oo..
.
oo.. (8 orientations)
..oo
o..o
.oo.
.
o.o. (4 orientations)
.o.o
.o.o
o.o.
.
o.o. (4 orientations)
.o.o
o..o
.oo.
.
.oo. (1 orientation)
o..o
o..o
.oo.

2012/13 British Mathematical Olympiad Round 1, problem 1.

cp's OEIS Frontend

User: Joseph Myers

A279766 Number of odd digits in the decimal expansions of integers 1 to n.

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A279259 Smallest positive integer m such that m, m+1, m+2, m+3 are divisible by 2n+1, 2n+3, 2n+5, 2n+7 respectively.

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A277608 Least number of fractions of the form (k+1)/k, for k a positive integer, whose product equals n.

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A264998 Number of partitions of n into distinct parts of the form 3^a*5^b or 2.

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Haskell

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A264997 Number of partitions of n into distinct parts of the form 3^a*5^b.

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Haskell

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A227715 Triangle read by rows: Number of 2n-step self-avoiding walks on diamond lattice ending at point with x = 2k.

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A227716 Triangle read by rows: Number of (2n+1)-step self-avoiding walks on diamond lattice ending at point with x = 2k+1.

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A227511 Triangle read by rows: Number of n-step self-avoiding walks on f.c.c. lattice ending at point with x = k.

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A227338 Number of n-step self-avoiding walks on cubic lattice ending at point with x = k.

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