A116416 -id:A116416 - cp's OEIS frontend

A272020 Irregular triangle read by rows: strictly decreasing sequences of positive numbers given in lexicographic order.

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

Offset: 0

Peter Kagey, Apr 17 2016

Length of n-th row given by A000120(n);
Min of n-th row given by A001511(n);
Sum of n-th row given by A029931(n);
Product of n-th row given by A096111(n);
Max of n-th row given by A113473(n);
Numerator of sum of reciprocals of n-th row given by A116416(n);
Denominator of sum of reciprocals of n-th row given by A116417(n);
LCM of n-th row given by A271410(n).
The first appearance of n is at A001787(n - 1).
n-th row begins at index A000788(n - 1) for n > 0.
Also the reversed positions of 1's in the reversed binary expansion of n. Also the reversed partial sums of the n-th composition in standard order (row n of A066099). Reversing rows gives A048793. - Gus Wiseman, Jan 17 2023

			Row n is given by the exponents in the binary expansion of 2*n. For example, row 5 = [3, 1] because 2*5 = 2^3 + 2^1.
Row 0: []
Row 1: [1]
Row 2: [2]
Row 3: [2, 1]
Row 4: [3]
Row 5: [3, 1]
Row 6: [3, 2]
Row 7: [3, 2, 1]

Peter Kagey, Table of n, a(n) for n = 0..10000

Cf. A000120, A096111, A116416, A116417, A271410, A001787.
Cf. A048793 gives the rows in reverse order.
Cf. A272011.
Lasts are A001511.
Heinz numbers of the rows are A019565.
Firsts are A029837 or A070939 or A113473.
Row sums are A029931.
A066099 lists standard comps, partial sums A358134, weighted sum A359042.
Cf. A005940, A059893, A125106, A161511, A242628.

T:= proc(n) local i, l, m; l:= NULL; m:= n;
      if n=0 then return [][] fi; for i while m>0 do
      if irem(m, 2, 'm')=1 then l:=i, l fi od; l
    end:
seq(T(n), n=0..35);  # Alois P. Heinz, Nov 27 2024

Table[Reverse[Join@@Position[Reverse[IntegerDigits[n,2]],1]],{n,0,100}] (* Gus Wiseman, Jan 17 2023 *)

A116417 If n = Sum_{m>=1} 2^(m-1) * b(n,m), where each b(n,m) is 0 or 1 and the sum is a finite sum, then a(n) = denominator of Sum_{m>=1} b(n,m)/m.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 6, 6, 4, 4, 4, 4, 12, 12, 12, 12, 5, 5, 10, 10, 15, 15, 30, 30, 20, 20, 20, 20, 60, 60, 60, 60, 6, 6, 3, 3, 2, 2, 1, 1, 12, 12, 12, 12, 4, 4, 4, 4, 30, 30, 15, 15, 10, 10, 5, 5, 60, 60, 60, 60, 20, 20, 20, 20, 7, 7, 14, 14, 21, 21, 42, 42, 28, 28, 28, 28, 84, 84, 84, 84

Offset: 0

Leroy Quet, Feb 13 2006

			13 in binary is 1101. So a(13) is the denominator of 1/4 + 1/3 + 1 = 19/12, since the binary digits at positions (from right to left) 1, 3 and 4 are each 1 and the other digits are 0.

Peter Kagey, Table of n, a(n) for n = 0..10000

Cf. A007088, A116416, A271410.

Table[Denominator@ Total@ MapIndexed[#1/ First@ #2 &, Reverse@ IntegerDigits[n, 2]], {n, 0, 79}] (* Michael De Vlieger, Aug 19 2017 *)

a(n) = {my(b = Vecrev(binary(n))); denominator(sum(k=1, #b, b[k]/k));} \\ Michel Marcus, Apr 18 2016

More terms from Joshua Zucker, May 03 2006

A272036 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is equal to 1.

Original entry on oeis.org

1, 38, 2090, 16902, 18954, 18988, 131334, 133386, 133420, 148258, 150284, 524314, 524348, 526386, 541212, 543250, 543284, 655644, 657682, 657716, 672568, 674580, 8388742, 8390794, 8390828, 8405666, 8407692, 8520098, 8522124, 8536962, 8536996, 8539048, 8913052, 8915090

Offset: 1

Michel Marcus, Apr 18 2016

nonn

That is, numbers such that both A116416(n) and A116417(n) are equal to 1.
Intersection of A272034 and A272035.
A number m with an exponent k in the binary sum must have another power of 2 having an exponent at least A275288(k). - David A. Corneth, Apr 01 2017

			For n=38, 2*38_10 = 2^6 + 2^3 + 2^2 = 1001100_2, and 1/2 + 1/3 + 1/6 = 1.

Robert Israel, Table of n, a(n) for n = 1..1655 (first 200 terms from Peter Kagey)

Cf. A116416, A116417, A272034, A272035, A272083, A275288.

Select[Range[2^20], Total[1/Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]] == 1 &] (* Michael De Vlieger, Apr 18 2016 *)

is(n) = my(b = Vecrev(binary(n))); sum(k=1, #b, b[k]/k) == 1;

A272034 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is the inverse of an integer.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 36, 38, 64, 128, 256, 512, 1024, 2048, 2056, 2080, 2088, 2090, 4096, 8192, 16384, 16896, 16900, 16902, 16928, 18944, 18952, 18954, 18988, 32768, 65536, 131072, 131328, 131332, 131334, 131360, 133376, 133384, 133386, 133420, 148224, 148256, 148258, 150284

Offset: 1

Michel Marcus, Apr 18 2016

nonn

That is, numbers such that A116416(n) is equal to 1.

The powers of 2 (A000079) form a subsequence.

			For n=36, 38_10=100100_2, and 1/3 + 1/6 = 1/2, the inverse of an integer.

Peter Kagey, Table of n, a(n) for n = 1..354

Cf. A000079, A116416, A116417, A272035, A272036, A272081.

Select[Range[2^18], IntegerQ[1/Total[1/# & /@ Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]]] &] (* Michael De Vlieger, Apr 18 2016 *)

isok(n) = {my(b = Vecrev(binary(n))); numerator(sum(k=1, #b, b[k]/k)) == 1;}

A272035 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is an integer.

Original entry on oeis.org

0, 1, 38, 39, 2090, 2091, 16902, 16903, 18954, 18955, 18988, 18989, 131334, 131335, 133386, 133387, 133420, 133421, 148258, 148259, 150284, 150285, 524314, 524315, 524348, 524349, 526386, 526387, 541212, 541213, 543250, 543251, 543284, 543285, 655644, 655645, 657682

Offset: 1

Michel Marcus, Apr 18 2016

nonn

That is, numbers such that A116416(n) equals 1.

2k is in this sequence if and only if 2k + 1 is. Therefore n + a(n) is odd for all n. - Peter Kagey, Apr 19 2016

			For n=39, 39_10=100111_2, and 1/1 + 1/2 + 1/3 + 1/6 = 2, an integer.

Peter Kagey, Table of n, a(n) for n = 1..450 (All terms less than 2^30)

Cf. A116416, A116417, A272034, A272036, A272082.

Select[Range[2^20], IntegerQ@ Total[1/Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]] &] (* Michael De Vlieger, Apr 18 2016 *)

isok(n) = {my(b = Vecrev(binary(n))); denominator(sum(k=1, #b, b[k]/k)) == 1;}

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A272020 Irregular triangle read by rows: strictly decreasing sequences of positive numbers given in lexicographic order.

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A116417 If n = Sum_{m>=1} 2^(m-1) * b(n,m), where each b(n,m) is 0 or 1 and the sum is a finite sum, then a(n) = denominator of Sum_{m>=1} b(n,m)/m.

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A272036 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is equal to 1.

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A272034 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is the inverse of an integer.

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A272035 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is an integer.

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