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A231007
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Number of months after which a date can fall on the same day of the week, but it is not possible that the two months have the same number of days, in the Gregorian calendar.
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2
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1, 11, 19, 27, 28, 44, 45, 53, 61, 70, 71, 73, 74, 83, 91, 99, 100, 116, 125, 131, 133, 143, 145, 146, 160, 171, 177, 185, 193, 202, 203, 205, 206, 215, 217, 223, 231, 232, 248, 249, 257, 263, 265, 274, 275, 277, 278, 287, 295, 303, 309, 320, 334, 335, 337, 347, 349, 355, 364
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OFFSET
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1,2
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COMMENTS
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In the Gregorian calendar, a non-century year is a leap year if and only if it is a multiple of 4 and a century year is a leap year if and only if it is a multiple of 400.
Assuming this fact, this sequence is periodic with a period of 4800.
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LINKS
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EXAMPLE
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1 belongs to this sequence because February 1, 2013 falls on the same day as March 1, 2013, but both February and March do not have the same number of days. In fact, a difference of 1 month can never produce the same calendar for the entire month, with the same number of days.
11 belongs to this sequence because December 1, 2011 falls on the same day as November 1, 2012 but both December and November do not have the same number of days. In fact, a difference of 11 months can never produce the same calendar for the entire month, with the same number of days.
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PROG
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(PARI) m=[0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5]; n=[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]; y=vector(4800, i, (m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-((i-1)\1200)+((i-1)\4800)-!((i-1)%48)+!((i-1)%1200)-!((i-1)%4800)-!((i-2)%48)+!((i-2)%1200)-!((i-2)%4800))))%7); x=vector(4800, i, n[((i-1)%12)+1]+!((i-2)%48)-!((i-2)%1200)+!((i-2)%4800)); for(p=0, 4800, j=0; for(q=0, 4800, if(y[(q%4800)+1]==y[((q+p)%4800)+1], j=1; break)); for(q=0, 4800, if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1], j=2; break)); if(j==1, print1(p", ")))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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