A083449 -id:A083449 - cp's OEIS frontend

A277833 Number of '3' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

0, 0, 1, 23, 349, 4721, 59553, 718985, 8424417, 96589849, 1089355281, 12128120713, 133626886145, 1459725651582, 15831824417065, 170663923183008, 1830096021953551, 19535528120770094, 207700960220046637, 2200466392323923180, 23239231824473799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '3' in the sequence 0, 1, 2, *3*, 4, ..., 12.
For n=3 there are 12 + 10 = 22 more digits '3' in { 13, 23, 30, ..., 39, 43, 53, ..., 123 }, where 33 accounts for two '3's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==3,digits(k)))))

A277833(n,m=3)=if(n>12, error("not yet implemented"), n>m, A277833(n,m+1)+(m+2)*10^(n-m-1), (9*n-11)*(10^n+1)\729+2-(m>n)) \\ M. F. Hasler, Nov 02 2016, edited Dec 29 2020

a(n) = A277832(n) - 4*10^(n-3) [for n >= 3] = A277834(n) + 5*10^(n-4) [for n >= 4].

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277835 Number of '5' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4665, 58993, 713385, 8368417, 96029849, 1083755281, 12072120713, 133066886145, 1454125651577, 15775824417009, 170103923182448, 1824496021947951, 19479528120714094, 207140960219486637, 2194866392318323180, 23183231824417799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n = 2 there is only one digit '5' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '5' in { 15, 25, ..., 45, 50, ..., 59, 65, ..., 115 }, where 55 accounts for two '5's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==5,digits(k)))))

A277835(n,m=5)=if(n>m,A277835(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 5,
a(5) = A277836(5) + 1, a(n) = A277836(n) + 7*10^(n-6) for n >= 6.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277836 Number of '6' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58986, 713315, 8367717, 96022849, 1083685281, 12071420713, 133059886145, 1454055651577, 15775124417009, 170096923182441, 1824426021947881, 19478828120713394, 207133960219479637, 2194796392318253180, 23182531824417099723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '6' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '6' in { 16, 26, ..., 56, 60, ..., 69, 76, 86, ..., 116 }, where 66 accounts for two '6's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

T[int_Integer, {bndsLow_Integer, bndsUpp_Integer}] := Table[
   Count[
    Flatten[Table[
      IntegerDigits[m],
      {m, 1, Sum[
         10^i - 1,
         {i, n}
         ]/9
       }
      ]],
    int
    ],
   {n, bndsLow, bndsUpp}
   ];
T[6, {0, 7}](* Robert P. P. McKone, Jan 01 2021 *)

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==6,digits(k)))))

A277836(n,m=6)=if(n>m,A277836(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277839(n) =  A083449(n) = A277830(n) - 1 for n < 6,
a(n) = A277835(n) - 7*10^(n-6) for n >= 6,
a(n) = A277837(n) + 8*10^(n-7) for n >= 7.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A211869 a(n) = Sum_{j=1..n-1} j(n-j)b^(j-1) with b = floor(n^2/4)+1.

Original entry on oeis.org

0, 1, 8, 98, 1712, 58985, 2541896, 187337236, 15687030920, 2014736789165, 280434300560320, 55591630021883014, 11642487182670742552, 3294318202343411333713, 969986091740868071844464, 371055858906757952457992360

Offset: 1

R. J. Cano, Feb 02 2013

Equivalently, a(n) is the number having the digits (j*(n-j); j=1..n-1), in base b = floor(n^2/4)+1.
From R. J. Cano, Mar 03 2018: (Start)
If a(n) were converted to the base 1+floor(n^2/4)=A033638(n) then a palindrome would be obtained. Such palindrome is related to A215940(n!);
a(7)=2541896 and A033638(7)=13, giving the palindrome "6ACCA6". Such palindrome cannot be converted directly to decimal, but it might be defined instead from these digits the polynomial f(t)= 6*t^5 +10*t^4 +12*t^3 +12*t^2+10*t^1+6*t^0, then evaluating for t=10, we get f(10)=713306=A215940(7!). 713306 clearly looks distinct than "6ACCA6". f(11) and f(12) respectively are 1130256 with "7021A6", and 1722942 with "6B10A6". Now evaluating f(14) we get 3646530 and if converted to base 14 it yields "6ACCA6". The same happens with f(15) converted to base 15, f(16) converted to Hexadecimal, and also in general for f(y) converted to base y, if it were provided that y>=13.
Here A033638(n) gives the lower bound for the infinite set of bases where this behavior can be observed. For simplicity it is chosen the base A033638(n) when defining this sequence, although what we actually want is to keep the pattern generated by the products j*(n-j). (End)
This sequence together with A033638 and A215940 demonstrates the connection among permutation sets and palindromes obtained by symmetric products. - Alexander R. Povolotsky, Feb 08 2013

			For n=5, the four products are 1*4 = 4, 2*3 = 6, 3*2 = 6, 4*1 = 4, giving the base-7 concatenation 4664. In base 10, this is a(5) = 1712.
For a(6) we have that 1+floor(6^2/4) = 10 so there is no need of converting the concatenation to decimal. By definition the products are j*(n-j) for j in 1..5: 1*(6-1) = 5 = 5*(6-5), 2*(6-2) = 8 = 4*(6-2), 3*(6-3) = 9 so the result is a(6)=58985.

R. J. Cano, Table of n, a(n) for n = 1..57
R. J. Cano, Additional information on this sequence.
R. J. Cano, A211869: Gauss taught me that.

Cf. A033638, A215940, A083449, A019566, A002113.

a(n,base=1+n^2\4)=sum(j=1, n-1, j*(n-j)*base^(n-1-j));

a(n) = Sum_{j=1..n-1} j*(n-j)*A033638(n)^(n-1-j).

A277834 Number of '4' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 344, 4671, 59053, 713985, 8374417, 96089849, 1084355281, 12078120713, 133126886145, 1454725651577, 15781824417015, 170163923182508, 1825096021948551, 19485528120720094, 207200960219546637, 2195466392318923180, 23189231824423799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '4' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '4' in { 14, 24, 34, 40, ..., 49, 54, ..., 114 }, where 44 accounts for two '4's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==4,digits(k)))))

A277834(n,m=4)=if(n>m,A277833(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n))

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 4,
a(n) = A277833(n) - 5*10^(n-4) for n >= 4, a(n) = A277835(n) + 6*10^(n-5) for n >= 5.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

cp's OEIS Frontend

A277833 Number of '3' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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A277835 Number of '5' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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A277836 Number of '6' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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Mathematica

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A211869 a(n) = Sum_{j=1..n-1} j*(n-j)*b^(j-1) with b = floor(n^2/4)+1.

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A277834 Number of '4' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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A211869 a(n) = Sum_{j=1..n-1} j(n-j)b^(j-1) with b = floor(n^2/4)+1.