A083575 -id:A083575 - cp's OEIS frontend

A103478 Positive integers k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.

5, 23, 73, 125, 1945, 23471

Offset: 1

Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005

On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 5 and with n exactly 2 greater than m, then that m belongs to this sequence. - Jeppe Stig Nielsen, Dec 04 2018

			a(1)=5 because 5 is the smallest positive integer k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.

Wilfrid Keller, Prime factors k*2^n + 1 of Fermat numbers F_m

Cf. A000215, A083575, A103477, A103479.

Select[Range[1, 2000], Mod[1 + PowerMod[2, 2^#, 1 + 5*2^(# + 2)], 1 + 5*2^(# + 2)] == 0 &] (* Julien Kluge, Jul 08 2016 *)

isok(n) = Mod(2, 1+5*2^(n+2))^(2^n) + 1 == 0; \\ Michel Marcus, Apr 29 2016

A199209 a(n) = 10*4^n+1.

Original entry on oeis.org

11, 41, 161, 641, 2561, 10241, 40961, 163841, 655361, 2621441, 10485761, 41943041, 167772161, 671088641, 2684354561, 10737418241, 42949672961, 171798691841, 687194767361, 2748779069441, 10995116277761, 43980465111041

Offset: 0

Vincenzo Librandi, Nov 04 2011

Bisection (odd part) of A083575. - Bruno Berselli, Nov 04 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A083575.

Magma
```
[10*4^n+1: n in [0..30]];
```

10*4^Range[0,30]+1 (* or *) LinearRecurrence[{5,-4},{11,41},30] (* Harvey P. Dale, May 27 2019 *)

a(n) = 4*a(n-1)-3.
a(n) = 5*a(n-1)-4*a(n-2).
G.f.: (11-14*x)/((1-x)*(1-4*x)).  - Bruno Berselli, Nov 04 2011

A248416 Rectangular array by antidiagonals: for n >= 0, row n gives the positions in the Thue-Morse sequence A010059 at which the first 2^n terms occur.

Original entry on oeis.org

1, 4, 1, 6, 4, 1, 7, 7, 7, 1, 10, 11, 13, 13, 1, 11, 13, 21, 25, 25, 1, 13, 16, 25, 41, 49, 49, 1, 16, 19, 31, 49, 81, 97, 97, 1, 18, 21, 37, 61, 97, 161, 193, 193, 1, 19, 25, 41, 73, 121, 193, 321, 385, 385, 1, 21, 28, 49, 81, 145, 241, 385, 641, 769, 769, 1, 24, 31, 55, 97, 161, 289, 481, 769, 1281, 1537, 1537, 1

Offset: 1

Clark Kimberling, Oct 06 2014

Each row contains contains its successor as a proper subsequence.
Note that this supposes that the Thue-Morse sequence A010059 has offset 1, whereas the true offset is 0. So really the entries should all be reduced by 1. - N. J. A. Sloane, Jul 01 2016
Apparently T(n,3) = A004119(n+1) for n>0. Apparently T(n,4) = A083575(n) for n>0. - R. J. Mathar, Nov 06 2018

			Northwest corner, n>=0, k>=1:
   1    4    6    7   10   11   13   16   18   19
   1    4    7   11   13   16   19   21   25   28
   1    7   13   21   25   31   37   41   49   55
   1   13   25   41   49   61   73   81   97  109
   1   25   49   81   97  121  145  161  193  217
   1   49   97  161  193  241  289  321  385  433
   1   97  193  321  385  481  577  641  769  865
The Thue-Morse sequence A010059 begins with 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, from which we see that the first 4 terms (=1,0,0,1) occur at positions 1, 7, 13, ..., as indicated for row n=2.

Cf. A010059 (Thue-Morse), A026147 (row 0), A091855 (row 1?), A157971 (row 2?),

Column 1 is essentially A004119 (or A181565).

A010060 := proc(n)
    local i;
    add(i, i=convert(n, base, 2)) mod 2 ;
end proc:
A010059 := proc(n)
    1-A010060(n) ;
end proc:
A248416Off0 := proc(n,k)
    option remember ;
    local strtN,binpat,src,thue ;
    if k = 1 then
        strtN := 0 ;
    else
        strtN := 1+procname(n,k-1) ;
    end if;
    binpat := [seq(A010059(i),i=0..n-1)] ;
    for src from strtN do
        thue := [seq(A010059(i),i=src..src+nops(binpat)-1)] ;
        if binpat=thue then
            return src ;
        end if;
    end do:
end proc:
A248416 := proc(n,k)
    1+A248416Off0(2^n,k) ;
end proc:
for d from 1 to 11 do
    for k from d to 1 by -1 do
        printf("%d,",A248416(d-k,k)) ;
end do: # R. J. Mathar, Nov 06 2018

z = 3000; u = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 20]; Length[u]
t[p_, q_] := t[p, q] = Table[u[[k]], {k, p, q}];
r[n_] := Select[Range[z], t[#, # + 2^(n - 1)] == t[1, 1 + 2^(n - 1)] &]
TableForm[Table[r[n], {n, 0, 10}]]

Definitions and examples clarified. - R. J. Mathar, Nov 06 2018

A288870 Triangle T from array A(k,n) = (2k+1)2^n + 1, k >=0, n >= 0 read by downwards antidiagonals.

Original entry on oeis.org

2, 3, 4, 5, 7, 6, 9, 13, 11, 8, 17, 25, 21, 15, 10, 33, 49, 41, 29, 19, 12, 65, 97, 81, 57, 37, 23, 14, 129, 193, 161, 113, 73, 45, 27, 16, 257, 385, 321, 225, 145, 89, 53, 31, 18, 513, 769, 641, 449, 289, 177, 105, 61, 35, 20, 1025, 1537, 1281, 897, 577, 353, 209, 121, 69, 39, 22

Offset: 0

Wolfdieter Lang, Jun 21 2017

This entry was motivated by a class work of Ferran D.

			The array A begins:
k\n  0  1  2   3   4   5    6    7    8    9    10 ...
0:   2  3  5   9  17  33   65  129  257  513  1025
1:   4  7 13  25  49  97  193  385  769 1537  3073
2:   6 11 21  41  81 161  321  641 1281 2561  5121
3:   8 15 29  57 113 225  449  897 1793 3585  7169
4:  10 19 37  73 145 289  577 1153 2305 4609  9217
5:  12 23 45  89 177 353  705 1409 2817 5633 11265
6:  14 27 53 105 209 417  833 1665 3329 6657 13313
7:  16 31 61 121 241 481  961 1921 3841 7681 15361
8:  18 35 69 137 273 545 1089 2177 4353 8705 17409
9:  20 39 77 153 305 609 1217 2433 4865 9729 19457
...
The triangle T begins:
m\k    0    1    2   3   4   5   6   7  8  9 10 ...
0:     2
1:     3    4
2:     5    7    6
3:     9   13   11   8
4:    17   25   21  15  10
5:    33   49   41  29  19  12
6:    65   97   81  57  37  23  14
7:   129  193  161 113  73  45  27 16
8:   257  385  321 225 145  89  53 31 18
9:   513  769  641 449 289 177 105 61 35 20
10: 1025 1537 1281 897 577 353 209 121 69 39 22
...

Cf. A288871. Columns of T (no 0's, or rows of A): A000051, A181565, A083575, A083686, A083705, A083683, A168596.

Row sums give A077802(n+1) or A095151(n+1).

Table[(2 k + 1)*2^(m - k) + 1, {m, 0, 10}, {k, 0, m}] // Flatten (* Michael De Vlieger, Jun 25 2017 *)

A(n, k) = (2*n + 1)*2^k + 1;
for(n=0, 10, for(k=0, n, print1(A(k, n - k),", "))) \\ Indranil Ghosh, Jun 22 2017

Array A(k, n) = (2*k+1)*2^n + 1 for k >= 0 and n >= 0.
Triangle T(m, k) = A(k, m-k) = (2*k+1)*2^(m-k) + 1, k >= m >= 0, otherwise T(m, k) = 0.
O.g.f. for column k of T: x^k*(2*(k+1) - (2*k+3)*x)/((1-2*x)*(1-x)), k >= 0.
E.g.f. for column k of T (without leading 0's): (2*k+1)*exp(2*x) + exp(x), k>=0.
E.g.f. for column k of T: 2^(-k)*(2*k+1)*exp(2*x) + exp(x) - S(k,x), with S(k, x) = 2^(-k)* Sum_{m=1..k} A288871(k,m)*x^(m-1)/(m-1)! if k >=1 and S(0,x) = 0.

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A103478 Positive integers k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.

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A199209 a(n) = 10*4^n+1.

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A248416 Rectangular array by antidiagonals: for n >= 0, row n gives the positions in the Thue-Morse sequence A010059 at which the first 2^n terms occur.

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A288870 Triangle T from array A(k,n) = (2*k+1)*2^n + 1, k >=0, n >= 0 read by downwards antidiagonals.

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Formula

A288870 Triangle T from array A(k,n) = (2k+1)2^n + 1, k >=0, n >= 0 read by downwards antidiagonals.